Alternative proofs to integral test that $\lim_{n\to\infty}\sum_{j=1}^n \alpha^{j^p}$ converges; $0<\alpha,p<1.$

Question

I can prove that $\large{\displaystyle\lim_{n\to\infty}\left(\sum_{j=1}^n \alpha^{\left(j^p\right)}\right)}$ converges, where $0<\alpha<1,\ $ and $\ 0<p<1,\ $ by proving that $$\large{\displaystyle\int_{x=1}^{\infty}\ \alpha^{\left(x^p\right)}} dx$$

converges by using substitution $\ x=u^{\frac{1}{p}},\ $ and then Integration by parts finitely many times. Then by the integral test, the original series must converge as well.

I am looking for alternative proofs to the integral test for the convergence of the original series. Can we use AM-GM, for example? Or some alternative comparison test?. Thanks in advance.

Answer 1

The Cauchy Condensation Test is a very powerful thing:

Assuming that $a_n\ge a_{n+1}\ge0$ we have $\sum a_n<\infty$ if and only if $\sum 2^na_{2^n}<\infty$.

It's actually easy to prove, or you can look it up. Apply it to $\sum n^{-p}$ and see what happens.

Here it's sufficient to show that $$\sum 2^n\alpha^{(2^n)^p}<\infty.$$Choose $k>0$ so $$\alpha^k<\frac12$$and then choose $N$ so $$2^{np}>kn\quad(n\ge N).$$Now $$\sum_{n\ge N}2^n\alpha^{2^{np}}\le\sum_{n\ge N}2^n\alpha^{kn}=\sum(2\alpha^k)^n<\infty.$$