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Questions tagged [telescopic-series]

For summation questions involving telescopic sums/series. This tag is often used with (summation) or (sequences-and-series).

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simplifying $ \frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + 2}=1 $ [duplicate]

$$ \frac{1}{\sqrt{2} + 1} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + 2}=1 $$ I came across this on a practice standardized test. The question was to evaluate the left hand side, and it ...
Keshinko's user avatar
  • 157
-2 votes
2 answers
63 views

AMC 10 Math Question About Formula For A Sequence... [closed]

Find the value of $ \frac{{1^2 + 1 \cdot 2 + 2^2}}{{1^3 \cdot 2^3}} + \frac{{2^2 + 2 \cdot 3 + 3^2}}{{2^3 \cdot 3^3}} + \cdots + \frac{{10^2 + 10 \cdot 11 + 11^2}}{{10^3 \cdot 11^3}} $ This is a ...
GalacticWood's user avatar
3 votes
0 answers
57 views

I need Help evaluate series :$\sum_{n=0}^{\infty} \frac{n!}{(x+1)(x+2)...(x+n+2)}$ [duplicate]

I need Help evaluate series :$$\sum_{n=0}^{\infty} \frac{n!}{(x+1)(x+2)...(x+n+2)}$$ $$s(x)=\sum_{n=0}^{\infty} \frac{n!}{(x+1)(x+2)...(x+n+2)}=\frac{1}{(x+1)(x+2)}+\frac{1}{x+1}\sum_{n=1}^{\infty}\...
Mostafa's user avatar
  • 2,288
2 votes
1 answer
132 views

Telescopic Sum in $\displaystyle \arcsin$ form

If $\displaystyle S_{n}$ $\displaystyle \sum^{n}_{k=1}\sin^{-1}\bigg[\frac{(2k+1)}{k(k+1)\sqrt{k(k+2)}+\sqrt{(k+1)(k-1)}}\bigg]$ Then $\displaystyle 100\cos(S_{99})=$ I am Trying to convert it into ...
jacky's user avatar
  • 5,172
1 vote
2 answers
91 views

Alternating sum involving binomial coefficients

I want to prove that $$ \sum_{i=0}^{n}{n\choose i} \frac{\left(1 + \alpha i\right)^{n} \left(-1\right)^{n - i}}{n!} = \alpha^{n}. $$ This is a guess based on the computations for $n = 0,1,2,3$. Do you ...
stackQandA's user avatar
-1 votes
1 answer
42 views

Problem with calculating sum of infinite series (maybe telescopic) [closed]

How can i calculate this: $\sum_{n=0}^{\infty}\frac{n+4}{(n+5)!}$ I already checked that it converges. I though about telescopic series but I'm unable to get there.
Dror's user avatar
  • 37
3 votes
1 answer
96 views

Find $\sum_{i=0}^n 7i(i+1)(i^2+i+1)^2$

$$\sum_{i=0}^n 7i(i+1)(i^2+i+1)^2$$ It just happens so that I knew this series is a telescopic with $$\sum_{i=0}^n (i+1)^7 - i^7 - 1$$ But how can I deduce from the top sequence that it is forming a ...
Samar's user avatar
  • 546
2 votes
0 answers
20 views

Are there any other sequences of functions besides $\prod_{j=0}^{k-1}(x+j) $ for which both their sum and the sum of their reciprocals telescope?

If $p_k(x) =\prod_{j=0}^{k-1}(x+j) $, then both $\sum_{j=a}^b p_k(x+j) $ and $\sum_{j=a}^b \dfrac1{p_k(x+j)} $ can be written as telescoping sums so we get $$\sum_{j=a}^b p_k(x+j) =\dfrac1{k+1}\left(...
marty cohen's user avatar
3 votes
1 answer
73 views

Solving the summation $ S = \sum_{r=1}^{25} \frac{1}{z^{18r} + z^{9r} + 1} $

Question: I am trying to solve the following summation problem involving complex numbers: Given a complex number $z$ satisfying $z^{26} = 1$, and the summation: $$ S = \sum_{r=1}^{25} \frac{1}{z^{18r} ...
OpateItZOpatoOpate's user avatar
3 votes
4 answers
83 views

Finding and proofing a closed formula for $\sum_{n=1}^k\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}$

I want to find and proof a closed formula for the following sum $$\sum_{n=1}^k\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\dots +\...
David Krell's user avatar
2 votes
1 answer
76 views

Show that $\sum_{k=4}^\infty \frac{9}{(k-1)^2} - \frac{9}{(k+2)^2} = \frac{769}{400}$

I'm trying to compute this series. $$S=\sum_{k=4}^\infty \dfrac{9}{(k-1)^2} - \dfrac{9}{(k+2)^2}$$ I have found the answer using a "shortcut" aka the Basel sum... $$S=\sum_{k=4}^\infty \...
LemNon's user avatar
  • 69
1 vote
1 answer
61 views

Find an efficient way to approximate the sum of the reciprocals of the squares of integers from 2017 to 4030

The question This question was taken from an International Junior Mathematical Olympiad exam for 9th graders. The question is as follows: Let $x$ satisfies the equation $\dfrac{1}{x}=\dfrac{1}{2017^2}...
Rahilly's user avatar
  • 11
7 votes
3 answers
304 views

Solved - Finding the function of $(1+x)(1+x^4)(1+x^{16})(1+x^{64})....$

The question states: For $0 < x < 1$, let $f(x) = (1+x)(1+x^4)(1+x^{16})(1+x^{64})(1+x^{256}).... $ Find $f(x).$ $f(x)= \prod_{n=0}^{\infty}(1+x^{(4^n)})$ I've tried noticing that it looks ...
thatbalanced's user avatar
2 votes
1 answer
91 views

Converge/Divergence of Telescoping Series

$\require{cancel}$ I'm trying to find whether the series $\sum_{n=1}^{\infty} \left(\frac{1}{\ln(n+2)} - \frac{1}{\ln(n+1)}\right)$ diverges or converges. It is a telescoping series. My approach was ...
Woshi's user avatar
  • 133
-1 votes
2 answers
128 views

Find the sum $\sum_{r=1}^n (r(r+1))^2 (-1)^r$ in terms of n.

Find the sum $$\sum_{r=1}^n (r(r+1))^2 (-1)^r$$ in terms of n. My approach I tried to make telescoping terms by adding another a constant term written in terms of r, $((r+2)-(r-1))$ but the $(-1)^r$ ...
Patrick Schick's user avatar
1 vote
1 answer
70 views

Calculate $\sum_{n=1}^\infty\frac{1}{(n-1)!+n!}$ [duplicate]

$$\text{Calculate } \sum_{n=1}^\infty\frac{1}{(n-1)!+n!}$$ I'm having a hard time solving this. I now I must create a minus to transform it into a telescopic series but I've been staring at it for a ...
J__n's user avatar
  • 1,123
0 votes
0 answers
95 views

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 \...
sagan's user avatar
  • 1
0 votes
0 answers
46 views

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 ...
Gabrielle Santos's user avatar
9 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 $\sum a_n$ ...
MSEU's user avatar
  • 1,146
0 votes
0 answers
43 views

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 ...
Lakshmi Priya's user avatar
2 votes
1 answer
108 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: $$\sum_{r = 1}^n \ln\left(\frac{r + 1}{r}\right) $$ ...
Tani's user avatar
  • 47
1 vote
1 answer
70 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^{...
Assandra Lakal's user avatar
-1 votes
1 answer
126 views

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?
Shub's user avatar
  • 596
1 vote
0 answers
34 views

$\sum_{k=1}^{+\infty} \ln_2(1 - (k+1)^{-2}) = 1$? [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}) = \...
mick's user avatar
  • 16.4k
0 votes
1 answer
65 views

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(...
Snowball's user avatar
  • 1,023
1 vote
2 answers
144 views

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 ...
shrey1's user avatar
  • 323
1 vote
1 answer
87 views

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(\...
Jonatan Iaczinski's user avatar
0 votes
0 answers
61 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)= \...
Jonatan Iaczinski's user avatar
-1 votes
1 answer
114 views

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 ...
Jonatan Iaczinski's user avatar
2 votes
3 answers
280 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}...
tyogi's user avatar
  • 305
2 votes
1 answer
73 views

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 $$\...
Jianxiang Zhu's user avatar
3 votes
1 answer
69 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 ...
John.Apple's user avatar
6 votes
4 answers
258 views

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 ...
Fadeel Khan's user avatar
0 votes
0 answers
144 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)^...
lopan's user avatar
  • 55
0 votes
1 answer
95 views

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 ...
Michael Jackson's user avatar
1 vote
1 answer
82 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’...
duckegg's user avatar
  • 669
-1 votes
1 answer
75 views

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 ...
math and physics forever's user avatar
0 votes
1 answer
31 views

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 ...
joeP's user avatar
  • 101
2 votes
1 answer
88 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 $...
Rishi Shekher's user avatar
4 votes
1 answer
99 views

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 ...
Aristarchus_'s user avatar
0 votes
5 answers
186 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}{\...
user112167's user avatar
2 votes
0 answers
113 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 $$\sum_{k=0}^n \binom{n}{k} = 2^n$$ ...
zhuli's user avatar
  • 2,571
2 votes
1 answer
78 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 ...
User's user avatar
  • 1,659
0 votes
1 answer
132 views

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^*|...
tonythestark's user avatar
0 votes
1 answer
92 views

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 ...
Z Ahmed's user avatar
  • 43.6k
1 vote
5 answers
157 views

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 ...
Shri's user avatar
  • 45
0 votes
0 answers
112 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} ...
Lai's user avatar
  • 22.2k
1 vote
0 answers
50 views

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 ...
spectr1's user avatar
  • 89
0 votes
0 answers
56 views

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 ...
fatehei's user avatar
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 ...
batchcoding____s's user avatar

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