Questions tagged [telescopic-series]
For summation questions involving telescopic sums/series. This tag is often used with (summation) or (sequences-and-series).
299
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Definite integral of ln(x) from 1 to a as the limit of a Riemann sum
For $a > 1$ determine the definite integral
$$
\int_1^a ln(x)
$$
as the limit of a Riemann sum. Hint: Use the partitioning $P_N = (x_0, x_1, …, x_N)$ with:
$$
1 = x_0 < x_1 < … < x_N = a \...
- 1
-4
votes
1
answer
80
views
how can i solve this question on telescoping series? [closed]
the integral part of the sum $$\sum_{n
=2}^{999} \frac{1}{n^{(2/3)}}$$ is ?
- 3
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0
answers
33
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Help for Telescopic Riemann sum
Consider the Riemann sum
$$\sum_{k=1}^n
2x^∗_k ∆x_k$$
of the integral of f(x) = 2x in an interval [a, b].
(a) Show that if $$x^∗_k$$ is the midpoint of the k−th subinterval, then the Riemann sum is
...
10
votes
1
answer
1k
views
Is every series a telescoping series?
This question may seem silly at first. We say that a series $\sum a_n$ is a telescoping series if there exists a sequence $(b_n)$ with $a_n=b_n-b_{n+1}$ for every $n$. One can show that $\sum a_n$ ...
- 900
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40
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1 + 1/2 * 1/3 + (1 * 3)/(2 * 4) * 1/5 + (1 * 3 * 5)/(2 * 4 * 6) * 1/7 + .. converges [duplicate]
To prove that $1+ \frac{1}{2}
\frac{1}{3}+\frac{1}{2} \frac{3}{4} \frac{1}{5} + \frac{1}{2} \frac{3}{4} \frac{5}{6}\frac{1}{7}+....
$ converges.
If i attempt to find the nth term for this series i ...
- 377
2
votes
1
answer
106
views
Does $\sum_{r = 1}^n \ln\left(\frac{1 + r}{r}\right) = \ln (\Gamma(n + 2)) - \ln (\Gamma(n+1))$? If so, why?
When attempting the evaluate the integral $\int_0^1 \{\ln(x)\}$, where $\{ x \}$ is the fractional part function, I came across the following sum:
$$\sum_{r = 1}^n \ln\left(\frac{r + 1}{r}\right) $$
...
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1
vote
1
answer
68
views
How to show $\frac{2}{2!}+\frac{7}{3!}+...+\frac{k^2-2}{k!}+...+\frac{9998}{100!}<3$ [duplicate]
Demonstrate the inequality$$\frac{2}{2!}+\frac{7}{3!}+...+\frac{k^2-2}{k!}+...+\frac{9998}{100!}<3$$
Attempt:
$$\sum^{100}_{n=2} \frac{n^2-2}{n!}$$
Note that $n^2-2 \leq n^2$, so we have:
$$\sum^{...
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-1
votes
1
answer
122
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Find the value $\sum_{n=a}^b\frac1{\sin (2^{n+3})}$ [closed]
Find the value of: $$\sum_{n=0}^{10}\frac1{\sin (2^{n+3})}$$
I'm stuck on this problem, can someone please help?
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0
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34
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$\sum_{k=1}^{+\infty} \ln_2(1 - (k+1)^{-2}) = 1$? [duplicate]
$$\sum_{k=1}^{+\infty} \ln_2(1 - (k+1)^{-2}) = 1$$
Where $\ln_2$ is the binary log(log base 2).
I assume this can be done by some sort of telescoping ?
$$\sum_{k=1}^{+\infty} \ln_2(1 - (k+1)^{-2}) = \...
- 15.3k
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1
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58
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Permuted sums of reciprocals
Let $\mathbb{S}_n$ be the set of all permutations of $[n]=\{1, \ldots, n\}$. For positive real numbers $d_1, \ldots, d_n$, prove
$$
\sum_{\sigma \in \mathbb{S}_n} \frac{1}{d_{\sigma(1)}\left(d_{\sigma(...
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1
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2
answers
136
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Summation of $\frac{1}{n(n+1)}$
I saw this problem initially and thought it would be pretty easy from some nice cancellation, but it just didn't happen:
$$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \ldots + \frac{1}{8010}$$
This can ...
- 323
1
vote
1
answer
80
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Evaluate $\sum_{n=1}^\infty \ln\left( \frac{n(2n+1)}{(n+1)(2n-1)}\right)$ [duplicate]
Evaluate $\sum_{n=1}^\infty \ln\left( \frac{n(2n+1)}{(n+1)(2n-1)}\right)$
My Approach
$$\sum_{n=1}^\infty \ln\left( \frac{n(2n+1)}{(n+1)(2n-1)}\right)=\sum_{n=1}^\infty ln(\left(n)(2n+1)\right) - ln(\...
0
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0
answers
55
views
Help with $\sum_{n=1}^\infty \ln\left( \frac{(n+1)^2}{n(n+2)} \right)$ [duplicate]
To what number the telescoping series converges,
$$\sum_{n=1}^\infty \ln\left( \frac{(n+1)^2}{n(n+2)} \right)$$
I already applied the log property of
$$\sum_{n=1}^\infty \ln((n+1)^2) - \ln(n)(n+2)= \...
-1
votes
1
answer
109
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The sum of $\frac{2}{(n-1)(n-2)(n)} $ converges to what number?
$$ \sum_{n=4}^\infty \frac{2}{(n-1)(n-2)(n)} $$ is a telescoping series but how can I make it more visual? How can I write that in telescoping form? $$\sum_{n=1}^\infty(a_n - a_{n+1})$$
I already ...
2
votes
3
answers
274
views
Creative telescoping of a non-hypergeometric function $\sum_{k=0}^n\binom{n}{k}p^k(1-p)^{n-k}A_{n,k}/\left[(1+A_{n,k})(b+cA_{n,k})(q+pA_{n,k})\right]$
I have been struggling with finding an analytic form for the following expression:
$$
a_n=\sum\limits_{k=0}^n\binom{n}{k}p^k(1-p)^{n-k}A_{n,k}\left[(1+A_{n,k})(1-q+(1-p)A_{n,k})(q+pA_{n,k})\right]^{-1}...
- 273
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1
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60
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Sum of the series created by a recursive sequence
$\{a_n\}$ and $\{b_n\}$ are two series of real number with $a_0=2$,$a_1=3$
and $a_{n+1}=3a_n-a_{n-1},\forall n\in \mathbb{N^+}$,when
$m= 2^n$, we have $\sqrt{b_n}=\frac{m}{a_m}$.Now, for the series
$$\...
3
votes
1
answer
68
views
On the limit defined by $A + B(A + B(A + B (A + B(\cdots))))$
Suppose $A$ and $B$ are some constant ($A,B\in\mathbb{R}$)
Is there a simple expression for $x$, where $x$ is:
$$
x=A+B[A+B[A+B[\cdots]]]]
$$
"..." indicates the pattern repeats forever.
In ...
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6
votes
4
answers
236
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Evaluate $\sum_{r=1}^{\infty} \dfrac{r^2 - 1}{r^4 + r^2 + 1}$
I was only able to observe that:
$\dfrac{r^2 - 1}{r^4 + r^2 + 1} = \dfrac{r^2 - 1}{(r^2 + r + 1)(r^2 - r + 1)}$
This hints at telescoping, but I would need an $r$ term in the numerator.
The original ...
- 374
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votes
0
answers
94
views
Spivak Calculus chapter 2 problem 6 - what's the intuition behind this technique?
The following technique could be used to derive the closed-form formulas for $1^a+...+n^a$ which I have verified so far for $a = 2,3,4$.
Let's assume $a=2$, expand $(k+1)^{a+1=3}$.
We now have $(k+1)^...
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votes
1
answer
86
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Find the value of $\sum^{\infty}_{n=0} \sum^{n}_{k=0} \frac{2^k}{2^n (k+1)(k+2)}.$
Find the value of $$
\sum^{\infty}_{n=0} \sum^{n}_{k=0} \frac{2^k}{2^n (k+1)(k+2)}.
$$
I'm guessing this has something to do with the cauchy product so my first thought was to dissect this series into ...
1
vote
1
answer
71
views
Telescoping a finite sum
Working through an upper-school pre-calculus book which starts with a bit of revision. Problem is that one of the questions in the development section of exercises is solvable using a technique they’...
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1
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75
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how to find the sum of these terms without the gamma function?
While solving a problem based on integration, I arrived at the following
$$\sum\limits_{x = 1}^{38} \ln\left(\frac{x}{x+1}\right)$$
I'm supposed to prove that this is less than $\ln(99)$
in order to ...
- 1,014
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1
answer
25
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Double EMA into Single EMA
The exponential moving average (EMA) operator is defined as:
$$y_t(x, \lambda) = (1-\lambda) \sum_{i=0}^\infty \lambda^i x_{t-i}$$
where $1-\lambda$ is the normalization factor, and the operator is a ...
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votes
1
answer
84
views
Best way to solve a summation with binomial coefficients in denominator apart from telecoping method
The value of $\sum_{r=1}^{m}\frac{(m+1)(r-1)m^{r-1}}{r\binom{m}{r}} = \lambda$ then the correct statement is/are
(1) If $m=15$ and $\lambda$ is divided by m then the remainder is 14.
(2) If $m=7$ and $...
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4
votes
1
answer
97
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Showing that the infinite series $\sum_1^\infty \left(\frac {1}{n} - \frac{1}{n+2}\right)$ is convergent
If we consider an infinite series witht the $n^{th}$ term $$a_n= \frac {1}{n} - \frac{1}{n+2}$$
for $n\ge1$
I am used to calculate the value to which a geometric series converges by looking at the ...
- 375
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votes
5
answers
172
views
How to calculate the sum $\sum_{i=1}^{\infty} \frac{6}{n(n+3)}$
This is a telescoping series whose sum is
$$\sum_{i=1}^{\infty} \frac{6}{n(n+3)} = \frac{11}{3}$$
I calculated it as $$\left(\frac{2}{1} - \frac{2}{1+3}\right) - \left(\frac{2}{\infty} - \frac{2}{\...
- 387
2
votes
0
answers
97
views
Using Gosper's algorithm to obtain the WZ certificate of $\sum \binom{n}{k} = 2^n$
I'm not sure where my work is wrong, I'm not obtaining an answer, even though I know there should be one.
In order to obtain the WZ proof certificate for the sum
$$\sum_{k=0}^n \binom{n}{k} = 2^n$$
...
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2
votes
1
answer
73
views
What is the value of $a_1a_2\cdots a_{2019}$?
Let $a_1=\frac 34$ and for any $n\geq2$ $4a_n=4a_{n-1}+\frac {2n+1}{1^3+2^3+\cdots n^3 }$. What is the value of $a_1a_2\cdots a_{2019}$?
I tried $1^3+2^3+\cdots +n^3=\frac {n^2(n+1)^2}{4}$ and
I ...
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1
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93
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Upper bound for Telescoping sum in gradient descent
I am studying a chapter in gradient descent . At some point we reach the sum in the left of the enequality and the writer says it's telescopic so this enequality holds:
$\sum_{t=1}^T \Big( ||x_t - x^*|...
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1
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82
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Summation of alternating series:$\sum_{k=1}^{n} (-1)^{k-1}k$
Alternate summations
$S_1=1-2+3-4+5-......+(2m-1)$ and $S_2=1-2+3-4+5-......-2m$
can be found as $\pm m$, respectively by arranging
$$S_1=[1+2+3+4+5+.....+(2m-1)]-4[1+2+3+4+....+m]$$
We can get the ...
- 42.9k
1
vote
5
answers
156
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Infinite sums of squares [closed]
$$\sum_{n=0}^{\infty} \frac {k^2(1-k)^2}{(n+k)^2(n+1-k)^2}$$
Here can anyone help me to solve this question,I can't think of any logic like telescopic, coefficient compare etc .
It would be helpful if ...
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0
answers
99
views
Is it a challenge to evaluate the indefinite integral $\int \frac{\sin n x}{\sin x} d x$, where $n\in N?$
Noting that
\begin{aligned}I_{k}-I_{k-2} &=\int \frac{\sin k x-\sin (k-2) x}{\sin x} d x \\&=2 \int \frac{\cos (k-1) x \sin x}{\sin x} d x \\&=2 \int \cos (k-1) x d x \\&=\frac{2}{k-1} ...
- 18k
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Need help in showing that the summation $\sum_{k=1}^{n} (a_{k+1}-a_{k}) = a_{n+1}-a_1$ [duplicate]
Given a sequence of real number $a_{1}$,$a_{2}$,...,$a_{n+1}$ show that $\sum_{k=1}^{n} (a_{k+1}-a_{k}) = a_{n+1}-a_1$
I am stuck on this problem we have been given by my lecturer. I don't have much ...
- 89
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votes
0
answers
55
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How to convert this into a telescopic sum [duplicate]
I'm trying to convert this series into a telescopic series but I'm stuck on making it
into a telescopic form by factoring it, tried partial faction decomposing it but couldn't proceed any further ...
- 1
11
votes
6
answers
1k
views
Alternative way to solve a limit problem
$$
\lim _{n \rightarrow \infty} \frac{1}{1+n^{2}}+\frac{2}{2+n^{2}}+\cdots+\frac{n}{n+n^{2}}
$$
I want to find the limit of this infinite series which I found in a book. The answer is $1/2$.
The ...
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1
vote
0
answers
57
views
Expressing $\sum\limits_{k=1}^{n-1}\frac{1}{k(k+1)}$ as $1 - \frac{1}{n}$
I was reading from INTRODUCTION TO ALGORITHM (THIRD EDITION) By Thomas H. Cormen and came across Telescoping series in the Appendix A (page 1148). And this was the definition:
For any sequence
$a_0,...
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votes
1
answer
115
views
Sum $ \sum_{n=1}^{\infty} {n2^n\over(n+2)!} $?
$$
\sum_{n=1}^{\infty} {n2^n\over(n+2)!}
$$
The exercise mentions that this can be written as a telescopic series;
I've been trying to write it in such a way but I'm stuck, can't seem to find one! Any ...
- 101
-1
votes
1
answer
98
views
Find the sum of series $\sum_{k=1}^\infty \frac{1}{k^2+2k}$ [duplicate]
Find the sum of the series.$$\sum_{k=1}^\infty \frac{1}{k^2+2k}$$
Which technique should I use? I tried but I cannot find anything.
- 13
2
votes
1
answer
94
views
Find the limit of $\frac{T_n}{5n+4}$
Given that $U_0=0$, $U_{n+1}=\frac{U_n+3}{5-U_n}$
Find the limit of $U_n$
Set $T_n= \sum_{k=1}^n \frac1{U_k-3}$, find $\lim_{n\to+\infty}\frac{T_n}{5n+4}$
Approaches
So for the first question, I ...
- 1,278
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votes
1
answer
81
views
Explicit formula for $S_n=\sum_{r=1}^n\dfrac1{r(r+1)}$. [duplicate]
I was asked to find an explicit formula for $$S_n=\sum_{r=1}^n\dfrac1{r(r+1)}$$ and then go on to find the limit.
I deduced that it would give $S_n=\frac1n-\frac1{n+1},$
however I was wrong and the ...
- 55
13
votes
1
answer
390
views
Proving $\sum_{n=1}^{99}\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}\lt\frac9{20}$
I found the original question asked by someone else, asking for this to be proven using only '9th grade math', this is the image:
Which can be written like
$$\sum_{n=1}^{99}\frac{\sqrt{n+1}-\sqrt{n}}{...
- 400
2
votes
5
answers
139
views
Value of $\sum_{n=0}^{1947}\frac{1}{2^n+\sqrt{2^{1947}}}$ $?$ [duplicate]
Find the value of $$\sum_{n=0}^{1947}\frac{1}{2^n+\sqrt{2^{1947}}}$$
MY APPROACH : I'm not able to telescope it . I tried to Rationalize it , but It was not possible . I've not attached any solution ...
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votes
1
answer
62
views
How to evaluate the following Product?
so I have the following product to evaluate :
$$P_{n}=\prod_{k=0}^{n-1} u_{k}$$
Where : $u_k = e^{w_n}$
All what I know about $u_k$ is that
$$
u_{n}=\frac{n+2}{n+1}
$$
So we'll have the following :
$$
...
-1
votes
3
answers
103
views
Summation of a finite sequence
This question is linked from my previous question: Summation of a sequence?
Given the sequence:
$$
a_n = 0.9^{n-1}a_1(1+d+d^2+...d^{n-1})
$$
and $a_1=100$ , $d= 1.5$
How to form an equation to find:
$$...
- 101
2
votes
3
answers
172
views
How to find if a series is telescopic
The series
$$\sum_{n=1}^\infty\left(\frac{4n+4}{3n+1}-\frac{4n}{3n-2}\right)$$
is telescopic and it converges to $-4+\dfrac43$.
But if we get the equivalent expresion
$$\sum_{n=1}^\infty\frac{-8}{9n^2-...
- 64.8k
3
votes
1
answer
160
views
Summation of powers of r and reciprocal of binomial coefficient
Evaluate the following sum: $$\sum_{r=1}^m\frac{(m+1)(r-1)(m^{r-1})}{r\binom{m}{r}}$$
where $\binom{m}{r}$ stands for ${}^mC_r$
I initially tried to change this into $$\frac{m+1}{m}\sum_{r=1}^m\frac{...
1
vote
1
answer
79
views
Prove that, $\sum_{k=1}^{n-1} \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}.$
The sequence $(a_k)_{k \geqslant 1}$ is an AP. Prove that, for every
$n \in \mathbb{Z^+}$, $$\sum_{k=1}^{n-1} \frac{1}{\sqrt{a_k} +
\sqrt{a_{k+1}}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}.$$
I ...
- 366
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votes
1
answer
48
views
Simplification of a telescopic series
I've read this question on Math SE, but I cannot figure out the following:
How to simplify $S$ to be $S=2-\frac{1}{a_{101}}$.
- 85
3
votes
0
answers
47
views
How to compute the limit of the infinite series $\sum_{n=1}^{\infty} \frac{1}{n(n+1)\ldots(n+m)}$, $m \in \mathbf{N}$ [duplicate]
For having some fun with infinite series, I am having a go at some of the exercises in Problems in Mathematical Analysis by Kaczor and Nowak. I would like to ask, ...
- 5,320
3
votes
1
answer
58
views
Find $\sum_{i=1}^n \frac{\sqrt{i+1}-\sqrt{i}}{\sqrt{i^2+i}}$
$$\sum_{i=1}^n \frac{\sqrt{i+1}-\sqrt{i}}{\sqrt{i^2+i}}$$
I have tried simplifying but I get $$\sum_{i=1}^n \frac{1}{\sqrt{i}}-\sum_{i=1}^n \frac{1}{\sqrt{i+1}}$$ which is subtraction of two ...
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