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I have this series:

$$\sum_{n=1}^\infty\frac{4}{(n+1)(n+2)}$$

and $$\sum_{n=1}^\infty\frac{4}{(n+1)(n+3)}$$

and I would like to know how to analyze, very step-by-step.

thanks

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First, apply partial fractions (if you don't know what that is, learn from Google). –  Ragib Zaman Feb 2 '12 at 14:34
    
Then, write the partial sum from 1 to $n$, and note that most of the terms cancel. –  David Mitra Feb 2 '12 at 14:35
    
I know how to make the partial fractions, I can't understand the next step. everywhere I search is not so easy for me to understand clearly –  Totty Feb 2 '12 at 14:47
    
The first series is in the revised version of the question "How to analyze convergence and sum of a telescopic series? I can't find a generic form" of yours. –  Américo Tavares Feb 2 '12 at 17:26

2 Answers 2

up vote 5 down vote accepted

One way to show convergence of the first series, is to first note that $$ \frac{4}{(n+1)(n+2)} < \frac{4}{(n+1)^2}$$ for all $n$. Then write $$ \sum_{n = 1}^{\infty} \frac{4}{(n+1)^2} = 4\sum_{n=2}^{\infty} \frac{1}{n^2}. $$ Presumably, you already know that the latter series converges, and so you can apply the comparison test, to show that $$ \sum_{n = 1}^{\infty} \frac{4}{(n+1)(n+2)} $$ converges. The other one can be treated in a similar way.

To find the sum, write $$ \frac{1}{(n+1)(n+2)} = \frac{A}{n+1} + \frac{B}{n+2},$$ and solve for $A$ and $B$ (this is called a partial fractions decomposition). Then you should be able to determine the $n$'th partial sum of this series explicitly, and easily evaluate the sum (these types of series will be called telescopic in your textbook, for obvious reasons). The other one can be treated in a similar way, but the computations will be a bit more complicated.

Note: Just realized that the first part of my answer is made redundant by the second, but I'll leave it - it can't hurt to know more than one way of accomplishing something.

Edit: After seeing your comment above, I'll add some more details on the next step too: $$ s_k = \sum_{n = 1}^{k} \frac{4}{(n+1)(n+2)} = 4\sum_{n = 1}^{k} \left( \frac{1}{n+1} - \frac{1}{n+2}\right).$$ Let us look more closely at the second sum $$\sum_{n = 1}^{k} \left( \frac{1}{n+1} - \frac{1}{n+2}\right) = \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \ldots = \frac{1}{2} - \frac{1}{k+2}.$$

So $s_k = 4(1/2 - 1/(k+2))$ and hence the sum of the series is $s = \lim_{k \to \infty} s_k = 4/2 = 2$.

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+1 for two arguments. A twice-proved theorem is twice as true. –  André Nicolas Feb 2 '12 at 15:19
    
thanks that helped!! :D –  Totty Feb 2 '12 at 15:56

For the second sum, take the terms $4\over (n+1)(n+3)$ and write them in the form $$\tag{1} {4\over (n+1)(n+3)}= {A\over n+1}+{B\over n+3}. $$ We need to figure out what $A$ and $B$ are. Towards that end, multiply both sides of the above by $(n+1)(n+3)$. This gives $$\tag{2} 4=A(n+3)+B(n+1). $$ We're presuming this holds for any value of $n$. If we set $n=-3$, then equation $(2)$ becomes $$ 4=B(-2); $$ which tells us $B=-2$.

Similarly, if we set $n=-1$ in equation $(2)$, we obtain $A=2$.

Thus, substituting these found values of $A$ and $B$ into equation $(1)$, we have $$\tag{3} {4\over (n+1)(n+3)}= {2\over n+1}-{2\over n+3}. $$

The second sum in your question can then be written as $$ \sum_{n=1}^\infty {4\over (n+1)(n+3)}= \sum_{n=1}^\infty \biggl( {2\over n+1}-{2\over n+3}\biggr). $$

Now, the infinite sum $\sum\limits_{n=1}^\infty a_n $ converges if and only if the sequence of partials sums $\{S_m\}$ defined by $S_m=\sum\limits_{n=1}^m a_n $ converges. In this case, we have $$ \sum\limits_{n=1}^\infty a_n = \lim_{m\rightarrow\infty}\sum\limits_{n=1}^m a_n. $$

Let's examine the $m^{\rm th}$ partial sum of our series. We explicitly write it out: $$ \eqalign{ \textstyle \sum\limits_{n=1}^m \Bigl[{2 \over n+1}\!\!-\!\! {2 \over n+3}\Bigr] &\!\! = \!\! \textstyle \Bigl[ ( \color{orange}{2\over 2} \!\! -\!\! \color{maroon}{2\over4} ) \!\! +\!\! (\color{orange} {2\over3}\!\! -\!\! \color{darkgreen}{2\over5} ) \!\! +\!\! (\color{maroon}{2\over4}\!\! - \!\! {2\over6})\!\! +\!\! (\color{darkgreen}{2\over5}\!\! - \!\! {2\over7})\!\! +\!\! \cdots \!\! + \!\! ({2\over (m-1)+1}\!\! -\!\! \color{pink}{2\over (m-1)+3} ) \!\! +\!\! ({2\over m+1}\!\! -\!\! \color{pink}{2\over m+3} ) \Bigr]\cr } $$ Note that the right most term in one of the parenthetical expressions cancels with the leftmost term of the second following parenthetical expression. After canceling everything, the only terms that survive are the $\color{orange}{\text{ leftmost terms of the first two parenthetical expressions}}$ and the $\color{pink}{\text{ rightmost terms of the last two parenthetical expressions}}$.

So $$ \eqalign{ \sum\limits_{n=1}^m \Bigl[{2 \over n+1}- {2 \over n+3}\Bigr] &= {1}+{2\over3}- {2\over (m-1)+3} -{2\over m+3} } $$

Now, taking the limit as $m\rightarrow\infty$ $$\lim_{m\rightarrow\infty} \sum\limits_{n=1}^m \Bigl[{2 \over n+1}- {2 \over n+3}\Bigr] =\lim_{m\rightarrow\infty}\Bigl[ 1+{2\over3}- {2\over (m-1)+3} -{2\over m+3}\Bigr] =1+{2\over3}. $$

It follows that $\sum\limits_{n=1}^\infty {4\over (n+1)(n+3)}=1+{2\over3}$.

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thanks this has helped me already.. (: –  Totty Feb 2 '12 at 15:56

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