# Decide whether the series are absolutely convergent , conditionally convergent, or divergent.

Decide whether the series are absolutely convergent, conditionally convergent, or divergent.

$$a) \sum _{n=1}^{\infty} (-1)^n \frac{(\log n)^{\log n}}{n^a} , a>0$$

$$b) \sum_{n=1}^{\infty} \frac {(-1)^{[\log n]}}{n}$$ where [ ] is greatest integer symbol.

My attempt : the easiest way to think about this problem is the Leibniz test both $a)$ and $b)$ will be conditionally convergents.

Is it correct ?

any hints/solutions

thanks u

For the second part note that for $n$ between $e^{k}$ and $e ^{k+1}$, $(-1)^{[\log n]}$ has a constant sign. Compare $\sum_{[e^{k}]}^{[e^{k}+1]} \frac 1 n$ with the integral of $\frac 1 x$ from $[e^{k}]$ to $[e^{k+1}]$ to see that the sum of the terms from $[e^{k}]$ to $[e^{k+1}]$ does not tend to $0$. Hence the series is divergent.

• sir how $e^k$ and$e^{k+1}$ come ? Aug 10, 2018 at 23:43
• @stupid If $[\ln n]=k$ then $k \leq \ln n < k+1$. This gives $e^{k} <n <e^{k+1}$. Aug 11, 2018 at 0:36
• sir thanks u... Aug 11, 2018 at 0:43

For the first one we have that

$$\frac{(\log n)^{\log n}}{n^a}\to \infty$$

indeed by $x=e^y \to \infty$ as $y\ge e^{2a}$

$$\frac{(\log x)^{\log x}}{x^a}=\frac{y^y}{e^{ay}}\ge\frac{e^{2ay}}{e^{ay}}=e^{ay}\to \infty$$

therefore the given series diverges.

For the second one we can write

$$\sum_{n=1}^{\infty} \frac {(-1)^{[\log n]}}{n}=\sum_{n=1}^{1} \frac {1}{n}-\sum_{n=2}^{2} \frac {1}{n}+\sum_{n=3}^{7} \frac {1}{n}-\sum_{n=8}^{20} \frac {1}{n}+\ldots+(-1)^{k+1}\sum_{n=[e^k]}^{[e^{k+1}-1]} \frac {1}{n}+\ldots =\sum_{k=1}^\infty (-1)^{k+1} a_k$$

that is an alternating sum of terms not convergent to zero and thus it diverges too, indeed as $k\to \infty$ by harmonic series

$$a_k=\sum_{n=[e^k]}^{[e^{k+1}-1]} \frac {1}{n}\sim \log (e^{k+1})-\log(e^k)=k+1-k=1$$

• @RushabhMehta Once $a_n \not \to 0$ the series diverges anyway, the$(-1)^n$ factor is completely useless to conclude here.
– user
Aug 10, 2018 at 22:17
• @RushabhMehta For the second one I'm thinking about it, it is not so trivial as the first one.
– user
Aug 10, 2018 at 22:18
• @RushabhMehta Now it should be complete. And please revise also your wrong first coment about the $(-1)^n$ term.
– user
Aug 10, 2018 at 22:44
• thanks u @gimusi..one last confusion how $y \ge e^{2a}$ ? Aug 10, 2018 at 23:51
• @stupid As $y \ge e^{2a}$ we have $(y)^y\ge (e^{2a})^y=e^{2ay}$. Why are you considering $\log(\log y)>2$?
– user
Aug 11, 2018 at 6:41