Determine whether $\sum_{n=1}^{\infty}\frac{n2^n}{n+3^n}$ converges.

Question

L'Hôspital is not allowed, we are allowed to use algebraic methods.

First I have tried the root-criteria:

$$\limsup_{n\rightarrow \infty}\sqrt[n]\frac{n2^n}{n+3^n}\iff \lim_{n\rightarrow \infty}\frac{2}{\sqrt[n]{n+3^n}}\,.$$ I don't know how to proceed further.

Then I have tried for a long time the quotient criteria, i.e.,

$$\limsup_{n\rightarrow \infty}\frac{\frac{(n+1)2^{n+1}}{n+1+3^{n+1}}}{\frac{n2^n}{n+3^n}}\iff 2\lim_{n\rightarrow \infty}\frac{n+3^n}{n+1+3^{n+1}}\,.$$

Last I have tried Discrete Comparison Test. My idea comparing

$$\frac{n2^n}{n+3^n}\text{ vs }\frac{1}{n}\,,\text{ i.e. },\frac{n^22^n}{n+3^n}\text{ vs }1\,.$$

I have seen the plots on WolframAlpha and I have seen that $n^22^n$ converges faster then $n+3^n$ therefore $\frac{n2^n}{n+3^n}>\frac{1}{n}$ for almost all $n$ therefore it diverges, but I am not sure whether one can argue like that. And how to prove that $n^22^n$ converges faster then $n+3^n$ formally.

Thanks for the support!

Answer 1

HINT:

Note that we have

$$0\le \frac{n2^n}{n+3^n}\le n\left(\frac23\right)^n$$

Can you finish now?

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If you want to apply the root test directly, then

$$\begin{align} \lim_{n\to \infty}\sqrt[n]{\frac{n2^n}{n+3^n}}&=\frac23\lim_{n\to \infty}\left(\frac{n^{1/n}}{\left(1+\frac n{3^n}\right)^{1/n}}\right)\\\\ &=\frac23\lim_{n\to \infty}\left( n^{1/n}e^{-\frac1n \log\left(1+\frac n{3^n}\right)}\right) \end{align}$$

Can you evaluate the limit $\lim_{n\to \infty}\frac{\log\left(1+\frac n{3^n}\right)}{n}$?

Answer 2

We may use the ratio test.

Consider $$|\dfrac{a_{n+1}}{a_n}|=\left|\dfrac{\dfrac{n2^{n+1}}{n+1+3^{n+1}}}{\dfrac{n2^n}{n+3^n}}\right|=\left|\dfrac{2(n+1)(n+3^n)}{n(n+1+3^{n+1}}\right|$$

Taking the limit as $n\to\infty$, we see it approaches $2/3 \lt 1$.

Therefore by ratio test, your series converges.

Answer 3

As an alternative by ratio test

$$\frac{(n+1)2^{n+1}}{n+1+3^{n+1}}\frac{3^n}{n2^n}=\frac{2(n+1)}{n}\frac{n+3^n}{n+1+3^{n+1}}\to \frac 23$$

and therefore the series converges.