0
$\begingroup$

I am looking into series and I am stuck at a point when I need to prove convergence of a series, without the convergence tests, and calculate the sum if it converges. I know that a geometric series ($\sum_{n=0}^\infty q^n$) converges when $ |q| \le 1 $ but I don't know what happens in the case of: $$ \sum_{n=1}^\infty \frac{1}{n(n+2)} $$ for example, or: $$ \sum_{n=1}^\infty \frac{n}{n+1} $$ Can you help me understand what is the actual partition of the series? Thank you

$\endgroup$
  • $\begingroup$ The second one blows up. The first is half of $\left(\frac{1}{1}-\frac{1}{3}\right)+ \left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+ \left(\frac{1}{4}-\frac{1}{6}\right)\cdots$. Almost everything cancels, and we get $\frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}\right)$. $\endgroup$ – André Nicolas Jan 21 '14 at 8:14
3
$\begingroup$

You know from basic analysis that a series converges when the $\lim\limits_{n \to \infty} \sum\limits_{i = 1}^n a_n$ converges (this is called the sequence of partial sums). In general if you are to actually show that a series converges to a value (which is actually pretty hard to do in general) you need to get some kind of expression for these partial sums and then just take a limit.

For the first series my recommendation would be to take a close look at the form in which it is presented to you and see if you can think of applying the partial fractions decomposition to it, and cancelling terms to get such an expression and then showing that this expression has a limit.

For the second series I would think about this: what is the limit of the terms of the series? That is, what is:

$$ \lim_{n \to \infty} \frac{n}{n+1}? $$

And in general, can a positive series (positive meaning all the terms are positive) converge if the terms are always greater than some $\epsilon > 0$?

$\endgroup$
1
$\begingroup$

The second sum blows up. For each term is $\ge \frac{1}{2}$.

The first sum is an instance of a telescoping sum. Note that $$\frac{1}{n(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right).$$ Thus the first sum is equal to
$$\frac{1}{2}\left[\left(\frac{1}{1}-\frac{1}{3}\right)+ \left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+ \left(\frac{1}{4}-\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\cdots\right].$$

Almost everything cancels, and we get $\frac{1}{2}\left[\frac{1}{1}+\frac{1}{2}\right]$.

Remark: For a more formal version of the solution to the first question, one should find $\sum_1^N \frac{1}{n(n+2)}$. After the cancellation, what will remain is $\frac{1}{2}\left[\frac{1}{1}+\frac{1}{2}\right]$ plus a couple of surviving terms that go to $0$ as $N\to\infty$.

$\endgroup$
0
$\begingroup$

For the first, if you decompose using partial fractions, you have
$$ \frac{1}{n(n+2)} = \frac{1}{2} (\frac{1}{n} - \frac{1}{n+2}) $$ and then, the terms telescope and the summation results in $$ \frac{1}{2}(1 + \frac{1}{2}) = \frac{3}{4} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.