I was reading about telescoping series and saw this example and was interested in its result. But I was a little confused because some of the steps to the derivation were skipped. In particular, I am interested in knowing why $r$ was removed from $u_{n}$  and how it was decomposed to yield $\frac{1}{k(k+1)...(k+r-1)} - \frac{1}{(k+1)...(k+r)}$. Some explanation as to how the final result was determined would also be greatly appreciated. Thank you!  $$\sum_{k=1}^nu_n=\sum_{k=1}^n \frac{1}{k(k+1)...(k+r)} = \frac{1}{r} \sum_{k=1}^n \left( \frac{1}{k(k+1)...(k+r-1)} - \frac{1}{(k+1)...(k+r)} \right) = \frac{1}{r} \left( \frac{1}{r!} - \frac{n!}{(n+r)!} \right).$$
123432
I have the following series  $$\sum_{k=1}^\infty \frac{(k-1)!}{(k+N)!},\quad\text{where }N \in \mathbb{N}. $$  I have found out that the series is equal to  $$\sum_{k=1}^\infty \biggl(\frac{1}{k}\cdot \frac{1}{k+1}\cdot _{...} \cdot \frac{1}{k+N}\biggr)$$  I also know that for $N=1$ we can use partial fraction expansion and we get a telescoping sum of $1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + ... + \frac{1}{k+1}$ which leaves $1$ if $k$ is heading to $\infty$.  Does anyone have any idea how I can go about doing this ?
121877
This function seemed to be pretty much straight forward, but my solution is incorrect. I have two questions: 1. Where did I make a mistake? 2. I learned that there are shortcuts for finding a series (substitution / multiplication / division / differentiation / integration of both sides). Is there something that I can apply here?  $$f(x)=\frac{1}{(1-x)^2}  \qquad  f(0)=1$$ $$f'(x)=\frac{2}{(1-x)^3}  \qquad  f'(0)=2$$  $$f''(x)=\frac{6}{(1-x)^4}  \qquad  f''(0)=6$$  $$f'''(x)=\frac{24}{(1-x)^5}  \qquad  f'''(0)=24$$  $$f''''(x)=\frac{120}{(1-x)^6}  \qquad  f''''(0)=120$$ The series is then $$1+\frac{2x}{2!}+\frac{6x^2}{3!}+\frac{24x^3}{4!}+\frac{120x^4}{5!}$$ and when simplified is  $$1+x+x^2+x^3+x^4$$ But the correct answer is  $$1+2x+3x^2+4x^3+5x^4 $$
109457
If $\displaystyle S_{n}=\frac{x}{x+1}+\frac{x^2}{(x+1)(x^2+1)}+\frac{x^{2^{2}}}{(x+1)(x^2+1)(x^{2^2}+1)}+...........+\frac{x^{2^{n}}}{(x+1)(x^2+1)...(x^{2^{n}}+1)}$ Then $\displaystyle \lim_{n\rightarrow \infty}S_{n} = \;,$ Where $x&gt;1$  $\bf{My\; Try::}$ First we will calculate $\bf{r^{th}}$ term of the sequence. So $$\displaystyle \bf{T_{r}} = \frac{x^{2^{r}}}{(x+1)(x^2+1)............(x^{2^{n}}+1)} = \frac{x^{2^{r}}(x-1)}{x^{2^{r+1}}-1}$$ So We get $$\displaystyle \bf{T_{r}} = \frac{x^{2^{r}}(x-1)}{(x^{2^r}-1)(x^{2^{r}}+1)}$$ Now I did not Understand How can I convert into Telescopic Sum. Help me Thanks
37847
Actually I have no idea how to begin. I tried to represent it as a product of two series or as a sum of two series but have no result with this. Any idea ?
38461
I am trying to calculate the following series:  $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}$$  and I managed to reduce it to this term  $$\sum_{n=1}^{\infty}(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2n+4})$$  And here I am stuck. I tried writing down a few partial sums but I can't see the pattern, $\frac{1}{2}-\frac{1}{2}+\frac{1}{5}+\frac{1}{4}-\frac{1}{3}+\frac{1}{8}+...$ I cant seem to find a closed formula that we can calculate for $S_n$  How should I go about solving this question
105419
How do you derive the formula for the sum of the series when $S_n = \sum_{j=1}^n j^2$?  The relationship $(n, S_n)$ can be determined by a polynomial in $n$. You are supposed to use finite differences to determine the degree and then use a calculator to derive the function. But how?
53658
Consider the series:  $$a) \sum_{n=1}^\infty \frac{a^n}{ \prod_{k=1}^n \ (1+\sin\frac{1}{2k})}$$ $$b) \sum_{n=1}^\infty \frac{a^n}{ \prod_{k=1}^n \ (1+\tan\frac{3}{2k})}$$  Showing that these two are convergent (and absolutely convergent) it's no big deal for  $a\in R^*$. But I couldn't figure out how to solve it when $a=1$. My guess is that the general term has to be brought to another form. And that is my question: how to prove convergence when $a=1$.  Thanks for helping with this!
80788
I do have somewhat of a reasoning for this.   $$S = 1 + x + x^{2} + x^{3} +.. + x^{n} $$   Denoting the derivative of $S$ as $T$   $$T = 1 + 2x + 3x^{2} + 4x^{3} +... + nx^{n-1}$$  $$xT = x + 2x^{2} + 3x^{3} +... + nx^{n}$$  Writing the derivative of this series as $V$  $$V = 1 + 2^{2}x + 3^{2}x^{2} +... + n^{2}x^{n-1}$$  $$xV = x + 2^{2}x^{2} + 3^{2}x^{2} +... + n^{2}x^{n}$$  Which is ${\sum{n^{2}x^{n}}}$  However, I am unable to start from $\sum{n^{2}}$and finish where I started. Can someone give me systematic steps on how to manipulate ${\sum{n^{2}x^{n}}}$into a generating function?   The answers are a little different from what I expected.  Let me explain with $\sum nx^{n}$. This function is not the series of any known function but with one slight manipulation, we can change it.  $$\sum x\frac {d x^{n}}{dx}$$ $$=x \frac{d}{dx} {\sum x^{n}} = x\frac {d}{dx} \frac{1}{1-x}$$ $$=\frac {x}{ (1-x)^2 }$$  I wanted a similar trick for evaluating $\sum n^{2}x^{n}$ $$=\sum nx \frac{d}{dx} x^{n}$$ And, then what ? How can I finish this ? I know how to start at $S$ and finish here, but I'm having trouble doing the algebra of starting with this sum and finishing where I started.
108482
How can the following be proved by induction?  $$\sum\limits_{i=1}^{n} \frac{1}{n+i} = \sum\limits_{i=1}^{n} \left(\frac{1}{2i-1} - \frac{1}{2i}\right)$$ I am out of ideas after practicing for a while:  $$\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots+\left(\frac{1}{2n-1}-\frac{1}{2n}\right) $$  Does this involve telescoping series?
108190
Question :   Find the sum to n terms of the series $\frac{1}{1.2.3}+\frac{3}{2.3.4}+\frac{5}{3.4.5}+\frac{7}{4.5.6}+\cdots$   What I have done :   nth term of numerator and denominator is $2r-1$ and $r(r+1)(r+2)$ respectively.  Therefore the nth term of given series is :   $\frac{2r-1}{r(r+1)(r+2)} =\frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+2}$ .....(1)  By using partial fraction :   and solving for A,B and C we get A = 1/2, B = -1, C =1/2  Putting the values of A,B and C in (1) we get :   $\frac{1}{2r}-\frac{1}{r+1}+\frac{1}{2(r+2)}$  But by putting $r =1,2,3, \cdots$  I am not getting the answer. Please guide how to solve this problem . Thanks.
41783
Whilst working out  the following summation emerged: $$\sum_{m=0}^n\frac 1{2^{2m}}\binom {2m}m$$ The is equivalent to  $$\begin{align} \sum_{m=0}^n \frac {(2m-1)!!}{2m!!}&amp;=\frac 12+\frac {1\cdot3}{2\cdot 4}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}+\cdots +\frac{1\cdot 3\cdot 5\cdot \cdots \cdot(2n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot 2n}\\ &amp;=\frac 12\left(1+\frac 34\left(1+\frac 56\left(1+\cdots \left(1+\frac {2n-1}{2n}\right)\right)\right)\right) \end{align}$$  and terms are the same as coefficients in the expansion of $(1-x)^{-1/2}$.  Once the solution  $$ \frac {n+1}{2^{2n+1}}\binom {2n+2}{n+1}$$ is known, the telescoping sum can be easily derived, i.e.    $$\frac 1{2^{2m}}\binom {2m}m=\frac {m+1}{2^{2(m+1)-1}}\binom {2(m+1)}{m+1}-\frac m{2^{2m-1}}\binom {2m}m$$  However, without knowing this a priori, how would we have approached this problem?
98404
Prove via induction that $\sum_{i=1}^n \frac{i}{(i+1)!}=1- \frac{1}{(n+1)!}$ Having a very difficult time with this proof, have done pages of work but I keep ending up with 1/(k+2).  Not sure when to apply the induction hypothesis and how to get the result $1- \frac{1}{(n+2)!}$.  Please help! thanks guys, youre the greatest!
111461
I need to use the identity $\frac{1}{k^2-1}=\frac{1}{2}(\frac{1}{k-1}-\frac{1}{k+1})$ to evaluate $\sum_{k=2}^n \frac{1}{k^2-1}$.  I am confused about how to begin this proof.  Also, how am I supposed to use the identity here?  Do I need to use induction?
51864
Consider, $$ S= \sum_{k=0}^{k=\infty} \frac{ k!}{x^{k}} (-1)^k$$ now this is, $$ S = ( 1 +D+D^2 +D^3...) ( \frac{1}{x})$$ using geometeric series $$ 1+D+D^2.. = \frac{1}{1-D}$$ So, $$ S= \frac{1}{1-D} \frac{1}{x}$$ $$ S = \frac{1}{x-1}$$ Therefore , for x&lt;1 $$ S= \sum_{k=0}^{k=\infty}(-1)^k \frac{ k!}{x^{k}}=\frac{1}{x-1}$$ Is this proof correct? Can I find an expression for l.h.s for x&gt;1 ?
98278
$$\sum_{r=0}^{k-1}4^r$$ Hi, I was wondering whether anyone could explain how to work this out. I know the end result is $\frac{4^k-1}{3}$, but I don't know why or how to get there.  Thank you :D
25491
I was wondering if there is also a closed expression for the series   $$\sum_{n=0}^{\infty} \binom{n+k}{k} x^n$$ where $|x|&lt;1.$  A few examples suggest that the answer is $\frac{1}{(1-x)^{k+1}}$ but I don't see how to show this.
73720
Find out the sum of the following infinite series $$\frac{3}{2^2(1)(2)} + \frac{4}{2^3(2)(3)} +\dots+\frac{r+2}{2^{r+1}(r)(r+1)}+\cdots $$ up to $r\to\infty$.  MY TRY:-  I tried to split $r+2$ as $[(r+1) +{(r+1)-r}]$ so that I can cancel one term from each terms in the numerator. Then I got an expression which was like Harmonic-Geometric series. But I could not do further any more after this.