# Find a sequence $a_n$ such that its sum converges, but the sum of its logarithm doesn't - Generalisation Help

This question came up in a past paper that I was doing, but it seems to be a fairly common, standard question.

Give an example of a sequence $a_n$ such that $\sum(a_n)$ converges, but $\sum(\log(a_n))$ does not converge. Give an example of a sequence $b_n$ such that $\sum(b_n)$ does not converge, but $\sum(\log(a_n))$ does.

Now, this particular question isn't all that hard (I just used $\frac{1}{n!}$ for the convergent series in both, and used the inverse function to get the non-convergent series). However, I got my answer through a bit of rational thinking, and maybe having seen similar questions before.

My question is, can any of the tests for convergence be used to do their job backwards, ie. starting with the fact that the summation of the series converges (and the summation of the log of the series does not converge)? Are there any other ways of doing this, other than having a bit of a think?

Thanks a lot!

• $a_n=\frac{1}{n^2}$ ? May 9, 2015 at 8:40
• Sorry, I think I wasn't clear. I know there a lot of different ways of answering this question, reciprocal triangular numbers are another I had though of. I'm asking if there's a way of finding a series $a_n$ where any specified function of it, eg. $ln(a_n)$ or $a_n^k$ etc. is known to be convergent, and any other specified function of it is known to be divergent. May 9, 2015 at 8:47

For $\sum \log(a_n)$ to converge, we should first have as a necessary condition $\log(a_n)$ converges to $0$ (by the limit test), and this only happens when $a_n$ converges to $1$ (and this, by turn, implies that $\sum a_n$ diverges). So, we limit our search to sequences converging to $1$.
Suppose that the sequence we are looking for takes the form: $a_n = 1 + b_n$, where $b_n$ converges to $0$, and is positive just like $a_n$ is positive. Then, $\log(a_n) \sim b_n$. Thus, you may choose any sequence $b_n$ of positive entries such that $\sum b_n$ converges.
Thus, any $(a_n)$ of positive entries, having the form $a_n = 1 + b_n$ where $b_n$ is of positive terms and $\sum b_n$ converges, will satisfy: $\sum a_n$ diverges and $\sum \log(a_n)$ converges.
Now, any convergent series $\sum a_n$ of positive terms will necessarily have $\sum \log(a_n)$ divergent because $\lim \log(a_n) \neq 0$.