convergence of infinite series $\sum_{n=1}^\infty \frac{x^n}{(1+x)(1+x^2)(1+x^3)\cdot\cdot\cdot (1+ x^n)}$

Question

I am reviewing for my final exam, and viewed this question:

Decide whether the following infinite sum is convergent for all $x >1$: $$\sum_{n=1}^\infty \frac{x^n}{(1+x)(1+x^2)(1+x^3)\cdot\cdot\cdot (1+ x^n)}$$

I don't even know how to approach this question. I have no idea what test should I try. Any help would be highly appreciated! Thanks

Answer 1

Denote $u_n(x)$ the general term of the series then by the ratio test we have

$$\frac{u_{n+1}(x)}{u_n(x)}=\frac{x}{1+x^{n+1}}\sim_\infty\frac1{x^n}\xrightarrow{n\to\infty}0<1$$ so the given series is convergent for $x>1$.

Answer 2

Hint: $$0<\frac{x^n}{(1+x)(1+x^2)(1+x^3)\cdot\cdot\cdot (1+ x^n)}\le 2^{1-n},$$because $\frac{1}{1+x^k}<\frac 12$ and $\frac{x^n}{1+x^n}<1$.

Answer 3

@Sami's approach is definitely the most direct, but here is another: $$ \frac{x^n}{(1+x)(1+x^2)\dotsc(1+x^{n-1})(1+x^n)} < \frac{x^n}{x^{1+2+\dotsc+n-1+n}} = \frac{x^n}{x^{n(n+1)/2}} = x^{-n(n-1)/2} $$

From here, there are plenty of ways to proceed -- one approach would be the root test, but a nicer approach would be to realize that for $n\geq3$, you also have $x^{-n(n-1)/2}<x^{-n}$, which allows you to use the geometric series test.