How Do You Solve This Alternating Series Involving Logarithms?

• A
• JHansen
In summary, the given series is bounded and can be evaluated to arbitrary precision by using the squeeze theorem and a numerical limit. The series converges and can be rewritten as a sum of two series, one with alternating signs and one with a specific pattern of terms.
JHansen
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.

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

Hey @jedishrfu

I agree with you, I evaluated it the same way but this exercise is actually different since we have different conditions on ##a_n##. You could be right, my understanding could be wrong, I'm not certain. That's a really good question and what I'm trying to figure out. If I know that I think I'm one step closer =)

jedishrfu said:
I found this reference:

https://socratic.org/questions/how-do-you-test-the-series-sigma-lnn-n-from-n-is-1-oo-for-convergence

I don't think your interpretation ##a_n## is valid though. I think they are not defined meaning you can't draw a conclusion instead I think there must be some generating function for ##a_n## .

How does the ##a_n = (2 mod 4)## fit in?
Hi again! I'm so sorry, I made a typo that could be very confusing, it should be "##a_n = 3 ## for n= 2 mod 4". It doesn't seem like I can edit the post. Sorry!

Would this help you understand the problem better? Because I'm still lost.

JHansen said:
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.
I think that the above is incorrect. Instead, I'm almost certain it should be ##a_n = 3## for ##n = 2 \mod 4## and ##a_n = -1## otherwise.

JHansen
Is there a way to split the series into the sum of two series?
- one series is for ##a_n=-1## for all terms
- the other series would be for the 2 mod 4 terms replacing n=2+4m

##\sum_{n=2}^{\infty} a_n \frac{ln(n)}{n}## = (-1)##\sum_{n=2}^{\infty} \frac{ln(n)}{n}## + (4) ##\sum_{m=0}^{\infty} \frac{ln(2+4m)}{2+4m}##

This is just a wild guess noting the curious nature of the ##a_n## definition

JHansen
jedishrfu said:
Is there a way to split the series into the sum of two series?
- one series is for ##a_n=-1## for all terms
- the other series would be for the 2 mod 4 terms replacing n=2+4m

##\sum_{n=2}^{\infty} a_n \frac{ln(n)}{n}## = (-1)##\sum_{n=2}^{\infty} \frac{ln(n)}{n}## + (4) ##\sum_{m=0}^{\infty} \frac{ln(2+4m)}{2+4m}##

This is just a wild guess noting the curious nature of the ##a_n## definition

You can't do this as it destroys convergence of the series: the sum ##\sum_{n=2}^\infty\frac{\ln(n)}{n}## diverges.

A simpler example is that ##1-1/2+1/3-1/4+1/5\ldots## converges (to ##\ln(2)##), but you can't break it apart as ##(1+1/3+1/5+\ldots)-(1/2+1/4+1/6+\ldots)## as both of these terms diverge.

AndreasC, JHansen and jedishrfu
Mark44 said:
I think that the above is incorrect. Instead, I'm almost certain it should be ##a_n = 3## for ##n = 2 \mod 4## and ##a_n = -1## otherwise.
You are correct! This is right!

Infrared said:
You can't do this as it destroys convergence of the series: the sum ##\sum_{n=2}^\infty\frac{\ln(n)}{n}## diverges.

A simpler example is that ##1-1/2+1/3-1/4+1/5\ldots## converges (to ##\ln(2)##), but you can't break it apart as ##(1+1/3+1/5+\ldots)-(1/2+1/4+1/6+\ldots)## as both of these terms diverge.
Do you have any ideas on how, if possible, to rewrite the series such that we can tackle it with some convergence test?

First write it as:
$$\sum_{n=2}^{\infty} a_n\frac{\ln(n)}{n};\quad a_n=\begin{cases}3 &\mbox{if}\; n \;\equiv 2\bmod 4 \\ -1 &\mbox{otherwise} \end{cases}$$
Not easy to evaluate these types of series.

First make an argument it converges:
Since the series is bounded and the four partial sums:
\begin{aligned} S_k&=\sum_{n=2}^k a_n\frac{\ln(n)}{n};\quad k\equiv 2\bmod 4 \\ S_{k+1}&=S_k-\frac{\ln(k+1)}{k+1}\\ S_{k+2}&=S_{k+1}-\frac{\ln(k+2)}{k+2} \\ S_{k+3}&=S_{k+2}-\frac{\ln(k+3)}{k+3} \\ \end{aligned}
approach one another as ##k\to\infty##, by the squeeze theorem, the sum converges. I think that is ok but not sure.

So that if it converges, we could as a matter of practicality, evaluate it to arbitrary precision. Here's a numerical limit to a precision of 20 digits:

Mathematica:
In[286]:= Needs["NumericalCalculus"]
myA[n_] := If[Mod[n, 4] == 2,
3
,
-1
];
nVal = NLimit[Sum[myA[n] Log[n]/n, {n, 2, k}], k -> \[Infinity],
WorkingPrecision -> 20]Out[288]= 0.480263097755934

Last edited:
John Greger
That looks
aheight said:
First write it as:
$$\sum_{n=2}^{\infty} a_n\frac{\ln(n)}{n};\quad a_n=\begin{cases}3 &\mbox{if}\; n \;\equiv 2\bmod 4 \\ -1 &\mbox{otherwise} \end{cases}$$
Not easy to evaluate these types of series.

First make an argument it converges:
Since the series is bounded and the four partial sums:
\begin{aligned} S_k&=\sum_{n=2}^k a_n\frac{\ln(n)}{n};\quad k\equiv 2\bmod 4 \\ S_{k+1}&=S_k-\frac{\ln(k+1)}{k+1}\\ S_{k+2}&=S_{k+1}-\frac{\ln(k+2)}{k+2} \\ S_{k+3}&=S_{k+2}-\frac{\ln(k+3)}{k+3} \\ \end{aligned}
approach one another as ##k\to\infty##, by the squeeze theorem, the sum converges. I think that is ok but not sure.

So that if it converges, we could as a matter of practicality, evaluate it to arbitrary precision. Here's a numerical limit to a precision of 20 digits:

Mathematica:
In[286]:= Needs["NumericalCalculus"]
myA[n_] := If[Mod[n, 4] == 2,
3
,
-1
];
nVal = NLimit[Sum[myA[n] Log[n]/n, {n, 2, k}], k -> \[Infinity],
WorkingPrecision -> 20]Out[288]= 0.480263097755934
That looks like ##ln^2(2)##

Actually, Mathematica returns a symbolic result:

Mathematica:
In[300]:= theVal = Sum[myA[n] Log[n]/n, {n, 2, \[Infinity]}]
N[theVal, 20]

Out[300]= 1/4 (9 Log[2]^2 - 3 EulerGamma Log[4] - 3 Log[4]^2 +
Log[256] + 2 StieltjesGamma[1] - StieltjesGamma[1, 3/4] -
StieltjesGamma[1, 5/4])

Out[301]= 0.48045301391820142467

Might be interesting to figure out how that's computed. Note this answer agrees only with the first 3 digits of NLimit result above. Not sure what's causing the discrepancy or which one is more correct.

JHansen
Just some closure in the matter: Found a thread at stackexchange which shows how to show sum is indeed ##\ln^2(2)##: Stack Exchange link

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