All Questions
7
questions
2
votes
3
answers
82
views
Compute $\sum_{n=1}^\infty (\frac34)^n \frac{7n+32}{n(n+2)}$
Question: Compute $$\sum_{n=1}^\infty \left(\frac34\right)^n \frac{7n+32}{n(n+2)}$$
I first did the partial fraction decomposition into:
$$\sum_{n=1}^\infty \left(\frac34\right)^n \frac{16}n - \sum_{...
- 157
2
votes
5
answers
211
views
Finding sum of the series $\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)}$
Find the sum: $$\sum_{r=1}^{n}\frac{1}{(r)(r+d)(r+2d)(r+3d)}$$
My method:
I tried to split it into partial fractions like: $\dfrac{A}{r}, \dfrac{B}{r+d}$ etc. Using this method, we have 4 equations in ...
- 671
1
vote
2
answers
98
views
If $S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}$, then calculate $14S$.
If $$S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}\,$$ find the value of $14S$.
The question can be simplified to:
Find $S=\sum\limits_{k=1}^n\,t_k$ if $t_n=\dfrac{n}{1+n^2+n^...
- 2,642
2
votes
4
answers
748
views
Finding the infinite Sum of a series: $\sum\frac1{n(n+1)(n+2)}$ [duplicate]
Find the infinite Sum of the series with general term $\frac{1}{n(n+1)(n+2)}$.
I decomposed the fraction upto this $1/(2n)-1/(n+1)+1/(2n+4)$. But I find no link about cancelling terms. So how should ...
- 51
-2
votes
1
answer
427
views
Calculate $S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}$. [duplicate]
Calculate $S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}$. I know I posted this question already but I want a more detailed answer. For example, how you got from one step to another using the partial fraction ...
- 47
2
votes
1
answer
187
views
Finding the partial sum of the series $\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}$
Hi asked the following question yesterday: Obtaining the sum of a series
Given the answers to that question by wj32, I am now trying to solve the following problem:
Consider the series
$$\sum_{n=1}^{...
- 727
1
vote
2
answers
97
views
Obtaining the sum of a series
Given the following example:
Obtain the Sum of the series
$$\frac{1}{(2)(4)}+\frac{1}{(4)(6)}+\frac{1}{(6)(8)}+...+\frac{1}{(2n)2(n+1)}=\sum_{k=1}^n{\frac{1}{4k(k+1)}}=\frac{1}{4}\sum_{k=1}^n{\frac{...
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