Find the infinite sum: $\frac{1}{(1)(2)}+\frac{1}{(3)(4)}+\frac{1}{(5)(6)}+\cdots$.   I thought of expanding it as $\frac{1}{1}-\frac{1}{2}+\frac{1}{3}- \cdots$, but it would yield nothing useful as this runs into the non-convergence issue.
5213
I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k}  = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series   $$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}  = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$  Does it converge?   If so, what is its sum?
83974
How do I simplify the following series $$1 + \frac{1}{1+2}+ \frac{1}{1+2+3}+ \frac{1}{1+2+3+4} + \frac{1}{1+2+3+4+5} + \cdot\cdot\cdot + \frac{1}{1+2+3+\cdot\cdot\cdot+n}$$  From what I see, each term is the inverse of the sum of $n$ natural numbers. Assuming there are $N$ terms in the given series, $$a_N = \frac{1}{\sum\limits_{n = 1}^{N} n} = \frac{2}{n(n+1)}$$ $$\Rightarrow \sum\limits_{n=1}^{N} a_N = \sum\limits_{n=1}^{N} \frac{2}{n(n+1)}$$ ... and I'm stuck.  I've never actually done this kind of problem before (am new to sequences &amp; series).  So, a clear and detailed explanation of how to go about it would be most appreciated.  PS- I do not know how to do a telescoping series!!
63488
Evaluate the sum  $$x+\frac{2}{3}x^3+\frac{2}{3}\cdot\frac{4}{5}x^5+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}x^7+\dots$$  Totally no idea. I think this series may related to the $\sin x$ series because of those missing even powers. Another way of writing this series:  $$\sum_{k=0}^{\infty}\frac{(2k)!!}{(2k+1)!!}x^{2k+1}.$$
121550
There is a couple of limits that I failed to find:  $$\lim_{n\to\infty}\frac 1 2 + \frac 1 4 + \frac 1 {8} + \cdots + \frac {1}{2^{n}}$$  and  $$\lim_{n\to\infty}1 - \frac 1 3 + \frac 1 9 - \frac 1 {27} + \cdots + \frac {(-1)^{n-1}}{3^{n-1}}$$  There is no problem to calculate a limit using Wolfram Alfa or something like that. But what I am interested in is the method, not just a concrete result.  So my questions are:   What should I do when I need a limit of infinite sum? (are there any rules of thumb?) What theorems or topics from calculus should I know to solve these problems better?   I am new to math and will appreciate any help. Thank you!
40315
$$\frac{1}{1 \cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1}$$  What I have so far in the induction is:  $$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}$$  But I'm not sure where to go from there . . .
35109
How to find the sum of the following series $$\frac{1}{5} - \frac{1\cdot 4}{5\cdot 10} + \frac{1\cdot 4\cdot 7}{5\cdot 10\cdot 15} - \dots\,.?$$  I have no idea. I have written the general term and tested its convergence by Gauss' test for convergence, but they are neither the question nor the answer.
41206
$$\frac{1}{1 \times3} + \frac{1}{2\times5}+\frac{1}{3\times7} + \frac{1}{4\times9}+\cdots $$  How to find sum of this series?  I tried this: its $n$th term will be = $\frac{1}{n}-\frac{2}{2n+1}$; after that  I am not able to solve this.
51705
How to calculate the sum of sequence $$\frac{1}{\binom{n}{1}}+\frac{1}{\binom{n}{2}}+\frac{1}{\binom{n}{3}}+\cdots+\frac{1}{\binom{n}{n}}=？$$ How about its limit?
69036
Here is an interesting infinite series. It would be great to see a method to evaluate it, if possible. I know it converges to a little less than 11/40  $\displaystyle\sum_{k=1}^{\infty}\frac{1}{4^{k}+k!}$  I could not think of any good identities to start this.  Thanks a million to those who can show how to evaluate it.   Maybe even in general, $\displaystyle\sum_{k=1}^{\infty}\frac{1}{x^{k}+k!}$, where $x\geq 1$
89894
Evaluate : $$1+\frac{2}{6}+\frac{2\cdot 5}{6\cdot 12}+\frac{2\cdot 5\cdot 8}{6\cdot12\cdot 18}+\ldots$$  thought: I have tried to manipulate the series so that the sum upto n terms , can be simplified. But I can't figure out any profitable result.
36528
How to find the sum of the following series:  $\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot$  Any hints.
40742
Find the limit of the following series:  $$ 1 - \frac{1}{4} + \frac{1}{6} - \frac{1}{9} + \frac{1}{11} - \frac{1}{14} + \cdot \cdot \cdot  $$  If i go the integration way all is fine for a while but then things become pretty ugly. I'm trying to find out if there is some easier way to follow.
27856
A previous problem had us solving $\sum_{n=0}^{\infty} \frac{1}{4^n}$ which I calculated to be $\frac{4}{3}$ using a bit of mathematical manipulation.   Wonderful. Thank you for all the prompt responses.  Could anyone suggest an alternate technique that does not involve differentiation?
104990
What is the limit of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $?  The $n$th summand is $\frac{1}{(2n)(2n + 1)(2n+2)} = \frac{1}{4} \frac{1}{n(2n+1)(n+1)}$.  I have tried expressing this as a telescoping sum, or as the limit of Riemann sums of a partition (the usual methods I normally try when doing this type of question- what are some other strategies?)
20763
Trying to find the sum of the following infinite series:  $$ \displaystyle\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{(2n-1)3^{n-1}}$$  Any ideas on how to find this sum?
123540
My teacher posed an infinite series question to me today and I'm not quite sure how to start to go about it. $$\sum_{r=1}^{\infty} \dfrac{2^{-r}}{r^{2}}$$ Any hints would be much appreciated.
9386
Can you please explain why $$ \sum_{k=1}^{\infty} \dfrac{k}{2^k} = \dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots = 2 $$  I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$