Finding the Limit of Partial Sums for a Series

[Дьявол]

How to find sum of some series?

For example I got:

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

All I know is that the condition that I need to find the sum of series is the sum of all partial (separate sums) and I know that the sum must be convergent.

So $$\lim_{n \rightarrow \infty}(x_n)=0$$.

In my case it is true.

So, $$\sum_{n=1}^{\infty}{\frac{1}{n(n+2)}}=\frac{1}{1*3}+\frac{1}{2*4}+\frac{1}{3*5}...$$

I need to find $$\lim_{n \rightarrow \infty}(X_n)$$.

But how will I find the limit of the partial sums?

Thanks in advance.

 

[mathman]

Trick: 1/[n(n+2)] = (1/2)(1/n - 1/(n+2)). Separate the sum into sum over odd n and sum over even n. For each sum, the series telescopes to the first positive term. The final result is then (1/2)(1 + 1/2) = 3/4.

 

[Дьявол]

Do you mean like: $$\lim_{n \rightarrow \infty}X_n=\lim_{n \rightarrow \infty}[\frac{1}{2}(1- \frac{1}{3})+\frac{1}{2}(\frac{1}{2}-\frac{1}{4})+\frac{1}{2}(\frac{1}{3}-\frac{1}{5})+\frac{1}{2}(\frac{1}{4}-\frac{1}{6})]$$

I see that some terms cancel out, but how did you get (1/2)(1 + 1/2) = 3/4 ?

 

[Office_Shredder]

You say some terms cancel... look at which ones don't cancel (not many)

 

[Дьявол]

Here cancel 1/3-1/3, 1/4-1/4.
1/2 * 1 for sure is not canceling, and also 1/2 * 1/2, -1/2 * 1/5, -1/2 * 1/6
As I can see 1/2*1/n+2 will always stay there. How to get in order these ones?

 

[mathman]

Do you mean like: $$\lim_{n \rightarrow \infty}X_n=\lim_{n \rightarrow \infty}[\frac{1}{2}(1- \frac{1}{3})+\frac{1}{2}(\frac{1}{2}-\frac{1}{4})+\frac{1}{2}(\frac{1}{3}-\frac{1}{5})+\frac{1}{2}(\frac{1}{4}-\frac{1}{6})]$$

I see that some terms cancel out, but how did you get (1/2)(1 + 1/2) = 3/4 ?

I said SEPARATE odd indices from even indices. You have the sum of two sums.

1/2( 1 - 1/3 + 1/3 - 1/5 + 1/5 -1/7 + ...) = 1/2

and 1/2(1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ...) = 1/4

Therefore total = 3/4

 

[Дьявол]

I said SEPARATE odd indices from even indices. You have the sum of two sums.

1/2( 1 - 1/3 + 1/3 - 1/5 + 1/5 -1/7 + ...) = 1/2

and 1/2(1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ...) = 1/4

Therefore total = 3/4

Oh, I see. Could you possibly tell me should I always use this method? I mean, should I always use $$\frac{x}{n}-\frac{y}{n+2}$$ depending from the denominator?

For example:

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

should I use $$\frac{1}{n^2}-\frac{n^2}{(n+1)^2}$$ ?

Thanks in advance.

 

[mathman]

Only if you do it right. (2n+1)/[n²(n+1)²] = 1/n² - 1/(n+1)²

 

[Дьявол]

Only if you do it right. (2n+1)/[n²(n+1)²] = 1/n² - 1/(n+1)²

Sorry, it was typo mistake.
$$\frac{2n+1}{n^2(n+1)^2}=\frac{A}{(n+1)^2}+\frac{2n+1}{n^2(n+1)^2} - \frac{A}{(n+1)^2}=\frac{A}{(n+1)^2}+\frac{2n+1-An^2}{n^2(n+1)^2}=$$
n=-1
$$2n+1-An^2=0$$
A=-1
$$=-\frac{1}{(n+1)^2}+\frac{(n+1)^2}{n^2(n+1)^2}=\frac{1}{n^2}-\frac{1}{(n+1)^2}$$

X_n=(1 - 1/4) + (1/4 - 1/9) + (1/9 - 1/16) +...+ (1/n² - 1/(n+1)²⁾

$$\lim_{n \rightarrow \infty}(X_n)=\lim_{n \rightarrow \infty}(1 - \frac{1}{(n+1)^2})=1-0=1$$

Thanks for the help and the efforts.