if we consider the series $\sum_{n\geq 1}{\frac{(-1)^{\lfloor| n-\ln n |\rfloor}}{n}}.$
It seems like an example of Leibniz, but here we have ${\lfloor | n-\ln n |\rfloor}$ which is annoying, how can we prove if it is convergent or divergent?
Asked
Active
Viewed 107 times
3
-
It might be treated similarly as this one: https://math.stackexchange.com/questions/1080849/is-the-series-sum-n-2-infty-frac-1-lfloor-sqrtn-rfloor-lnn-co?rq=1 – Sungjin Kim May 19 '17 at 14:54
-
yes, but here the problem here is $|ln(n)-n|$ – ron macguire May 19 '17 at 14:58
-
Do you know the proof of Dirichlet's criterion? – Daniel Fischer May 19 '17 at 15:00
-
yes, using abel's transformation (summation by parts) – ron macguire May 19 '17 at 15:01
-
Then you'll know that the boundedness of the partial sums is a stronger condition than needed. Here, look at the order of growth of $$\sum_{n = 1}^N (-1)^{\lfloor n - \ln n\rfloor}.$$ – Daniel Fischer May 19 '17 at 15:07
-
inside the floor there is absolute $|n-ln(n)|,$ or it doesn't have any sense, shouldn't we study the parity of $⌊|n−ln(n)|⌋$ – ron macguire May 19 '17 at 15:56
-
@ronmacguire Why should there be absolute value, is there any $n>1$ such that $n-ln(n)<0$ – Rab May 19 '17 at 15:59
-
Since $n > \ln n$ for all positive integers, the absolute value is unnecessary. – Daniel Fischer May 19 '17 at 15:59
1 Answers
3
The sum is equal to $$ \sum_{k=0}(-1)^k\sum_{e^{k-1}\lt n\le e^k}\frac{(-1)^n}{n}\tag{1} $$ By the Alternating Series Test, $$ \left|\,\sum_{e^{k-1}\lt n\le e^k}\frac{(-1)^n}{n}\,\right|\le e^{1-k}\tag{2} $$ Thus, by Comparison with a Geometric Series, the outer alternating sum in $(1)$ converges absolutely.
robjohn
- 345,667