Suppose $\displaystyle a_n=\frac{1}{(\text{log}2)^p+(\text{log3})^p+...+(\text{log}n)^p}$ for $n\geq 2$ and $p>0$. For which values of $p$ does $\sum_{n=2}^\infty a_n$ converge?

So, when $p\leq 1$, $\displaystyle a_n\geq \frac{1}{(n-1)(\text{log}n)^p}$ for $n\geq 2$. Let $\displaystyle b_n=\frac{1}{(n-1)(\text{log}n)^p}$. Then $\sum b_n$ diverges since $p\leq 1$. Therefore by comparison test $\sum a_n$ diverges when $p\leq 1$.

When $p>1$, $\displaystyle a_n\leq \frac{1}{(\text{log}n)^p}$. Let $\displaystyle c_n= \frac{1}{(\text{log}n)^p}$. If $\sum c_n$ converges then I will win. But now I am stuck. Is $\sum c_n$ converges? Then how can I prove it? If not how can I conclude the state of convergence of $\sum a_n$ for $p>1$?

Can anybody please help?

  • For $p\leq 1$ and $n>2$ we have $a_n\geq 1/n(\ln n)^p>1/n\ln n.$ – DanielWainfleet Aug 19 at 17:02

The series $\sum \frac1{\log^p n}$ diverges...

For $p>1$, note that $$a_n=\frac1{\sum_{k=2}^n \log^p k} \leq \frac1{\sum_{k=n/2}^n \log^p k} \leq \frac1{(n/3) \log^p (n/2)}$$ since the denominator contains at least $n/3$ terms, each of which at least $\log^p (n/2)$. Thus $$\sum a_n\leq \sum\frac3{n\log^p(n/2)}...$$

  • Good. But your first sentence seems irrelevant. – DanielWainfleet Aug 19 at 16:58

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