Prove that series $\sum \frac{a^n}{n^b(\log n)^c}$ is convergent for $|a|<1,b,c \in \mathbb{R}$ Prove that series $$\sum \frac{a^n}{n^b(\log n)^c}$$ is convergent for $|a|&lt;1,b,c \in \mathbb{R}$

when $b,c &gt;0$  Then $\ \ \frac{|a|^n}{n^b(\log n)^c} &lt; |a|^n$ and hence series is  converge by comparison test.

but how to prove for other values of $b$ and $c$. Any hint ??
$\sum{\frac{a^n}{n^b(\log n)^c}}&lt;\sum {\frac{a^n}{n^b(n)^c}}$
since

$a&lt;1$ we can observe that the graph of $f(x)={\frac{a^x}{x^{b+c}}}$ is similar to that of $f(x)=\frac{1}{x}$ and hence we can approximate the sum as the integral 

$\int_1^\infty\frac{a^x}{x^{b+c}}$ &lt;$\int_0^\infty \frac{a^x}{x^{b+c}}$

which is convergent  because it is basically the Gamma function Of (b+c+1) multiplied by some constant and although the gamma function does not converge for values less than 0, our integral does because it does not take values between (0,1).
$\sum{\frac{a^n}{n^b(\log n)^c}}&lt;\sum {\frac{a^n}{n^b(n)^c}}$
since

$a&lt;1$ we can observe that the graph of $f(x)={\frac{a^x}{x^{b+c}}}$ is similar to that of $f(x)=\frac{1}{x}$ and hence we can approximate the sum as the integral 

$\int_1^\infty\frac{a^x}{x^{b+c}}$ &lt;$\int_0^\infty \frac{a^x}{x^{b+c}}$

which is convergent  because it is basically the Gamma function Of (b+c+1) multiplied by some constant and although the gamma function does not converge for values less than 0, our integral does because it does not take values between (0,1).