- #1

JHansen

- 8

- 1

- TL;DR Summary
- Would like to know how to evaluate this non-trivial series :)

Hi!

Some time ago I came across a series and never solved it, I tried to give a new go because I was genuinely curious how to tackle it, which I thought would work, because it looks innocent, but there is something about the beast making it hard to approach for me. So need some help! Maybe this is straight forward for mathematicians, but not for me.

The series reads

$$\sum_{n=2}^{\infty} a_n \frac{ln(n)}{n}$$

Doesn't look that interesting right, at first glimpse one sees it will diverge. But now comes the problem (for me):

##a_n = 3## for ##a_n =## 2 mod 4 and ##a_n = -1## otherwise.

So I think the notation 2 mod 4 means that ##a_n = 3## for n=2,6,10,... so the summation constant will go like (3 -1 -1 -1 +3 -1 -1 -1 + 3 + ...) (starting from n=2).

But giving all the above I don't know how to evaluate the series, I have tried to re-write it without any progress. If you can help me tame this beast, you will help to make peace in my mind!

Some time ago I came across a series and never solved it, I tried to give a new go because I was genuinely curious how to tackle it, which I thought would work, because it looks innocent, but there is something about the beast making it hard to approach for me. So need some help! Maybe this is straight forward for mathematicians, but not for me.

The series reads

$$\sum_{n=2}^{\infty} a_n \frac{ln(n)}{n}$$

Doesn't look that interesting right, at first glimpse one sees it will diverge. But now comes the problem (for me):

##a_n = 3## for ##a_n =## 2 mod 4 and ##a_n = -1## otherwise.

So I think the notation 2 mod 4 means that ##a_n = 3## for n=2,6,10,... so the summation constant will go like (3 -1 -1 -1 +3 -1 -1 -1 + 3 + ...) (starting from n=2).

But giving all the above I don't know how to evaluate the series, I have tried to re-write it without any progress. If you can help me tame this beast, you will help to make peace in my mind!