Convergence of particular type of series

Question

I know that a series in the form

$$\sum_{n=1}^\infty\frac1{n^p} \tag 1$$

converges if $p>1$. But suppose I had a series in this form:

$$\sum_{n=1}^\infty\frac1{n^{p(n)}} \tag 2$$

where $p(n)$ is a function which is strictly greater than $1$ for each $n$ (but whose infimum might be $1$). Does the series still converge? Can I apply $(1)$ in this situation? An example of series of this kind may be

$$\sum_{n=1}^\infty\frac1{n^{2-\sin n}} \tag 3$$

How do you prove the convergence of $(3)$?

Does this answer your question? Convergence or divergence of $\sum_{n=1}^{\infty}\frac{1}{n^{2-\cos n}}$ See also https://math.stackexchange.com/questions/310756 and [related posts](https://approach0.xyz/search/?q=AND site%3Amath.stackexchange.com%2C OR content%3A%24%5Csum_%7Bn%3D1%7D%5E%5Cinfty%5Cfrac1%7Bn%5E%7B2-%5Csin n%7D%7D %24&p=1) — Anne Bauval, Jun 18 '24 at 14:23

score 1 · Answer 1 · answered Jun 18 '24 at 17:31

The answer simply depends on $p(n)$. For example, we know that the "logarithmic $p$-series"

$$\sum_{n \ge 2} \frac{1}{n (\ln n)^p}$$

also converges iff $p > 1$ (by the integral test, or Cauchy condensation). This gives that if

$$p(n) = \log_n \left( n (\ln n)^p \right) = 1 + \frac{p \ln \ln n}{\ln n}$$

(with an arbitrary modification for $n = 1$) then $\sum \frac{1}{n^{p(n)}}$ converges iff $p > 1$.

Convergence of particular type of series

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