# Determining whether the infinite series is convergent or divergent

Determine whether or not the following series converges, showing all working:$$\sum^\infty_{n=1} \frac{n^2}{3^n-2^n}$$

Working:

So I used the ratio test to begin and then split the equation into two sections

$\lim \limits_{n \to \infty} \vert \frac {n+1^2}{n^2}\vert$ $\lim \limits_{n \to \infty} \vert \frac {3^n-2^n}{3^{n+1}- 2^{n+1}}\vert$

I was then able to simplify the left limit and got it to be equal to $1$. I am however stuck on how I can simplify the limit $\lim \limits_{n \to \infty} \vert \frac {3^n-2^n}{3^{n+1}- 2^{n+1}}\vert$ into something which I can then determine what value it equals to. Any advice is greatly appreciated, really struggling with infinite series. Please feel free to edit my question for clarity. Thank you

• Hint: Write $3^n-2^n=3^n(1-(2/3)^n)$, and the same for $3^{n+1}-2^{n+1}$ – Kelenner Oct 27 '15 at 9:45
• Thank you for the quick reply. We I tried it this way I ended up with the answer being $1/2$, is that correct or have I done it wrong? – Joe Speedmen Oct 29 '15 at 8:16
• I think that the limit of $\frac{3^n-2^n}{3^{n+1}-2^{n+1}}$ is $1/3$, not $1/2$. A misprint ? – Kelenner Oct 29 '15 at 8:45
• Nah, I must have made an error somewhere, is $3^{n+1}- 2^{n+1}$ = $3^{n+1}(1- (2/3)^{n+1})$ equivalent to $(3)3^{n}(1- (2/3)(2/3)^{n})$, cause that might be where I have made my mistake – Joe Speedmen Oct 29 '15 at 8:51

Simple with asymptotic analysis: $$3^n-2^n\sim_\infty 3^n,\quad\text{hence} \quad \frac{n^2}{3^n-2^n}\sim_\infty\frac{n^2}{3^n},$$ and the latter converges since it is $o\Bigl(\dfrac1{n^2}\Bigr)$.

∑((n²)/(3ⁿ-2²)) is convergent because (1/(3ⁿ-2ⁿ))=(1/(3ⁿ(1-(2/3)ⁿ))) is comparable to 1/3ⁿ and so ∑((n²)/(3ⁿ-2²)) is comparable to ∑n²/3ⁿ and this is convergent by root test.

• Please try to use MathJax. There is a tutorial on Meta. – wythagoras Oct 27 '15 at 9:54