# 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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### 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?
1 vote
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1 vote
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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 ...
1 vote
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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 ...
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### 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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### 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 ...
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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 ...
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### 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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1 vote
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### 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 ...
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### 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 $S=2-\frac{1}{a_{101}}$.
### 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 ... 