# Confused as to which test to use to determine if series converges or diverges

For the question $$\sum^\infty_{n=0}\frac{n^2}{2^n+1}$$I first tried the root test as the denominator was a number to the power of n, but it would result in the numerator having a power to the n so I scrapped that idea. I tried to use the divergence test and then l'hopital's rule as both numerator and denominator went to infinity but I feel like that's over-complicating the question and that there's an easier test for it. Please help.

• Bound the sum using $\dfrac{n^2}{2^n}$ – MATHS MOD Mar 19 at 2:48
• The root test is the right way to go here. Look again at the numerator; you end up with essentially $n^{1/n}$. – Clayton Mar 19 at 2:48
• Use Cauchy condensation test – MATHS MOD Mar 19 at 2:50
• Okay so I bound the sum using MATHSMOD suggestion and took the root test from Clayton but the cauchy condensation test is not part of the topic so I'm not allowed to use it yet. So do I just find $\lim_{n->\infty}\frac{n^{\frac{2}{n}}}{2}?$ – Random Student Mar 19 at 2:55
• Yes you should find the limit. The easiest way to do this is to find the logarithm of the limit, and then take the exponential of your result. – Spencer Mar 19 at 2:57

You can combine the comparison test $$\sum \frac{n^2}{2^n+1} < \sum \frac{n^2}{2^n}$$ with the ratio test, since $$\lim_{n\to\infty} \bigg| \frac{(n+1)^2/2^{n+1}}{n^2/2^n} \bigg| = \lim_{n\to\infty} \frac{n^2+2n+1}{2n^2} = \frac12,$$ indicating convergence. You can also use the ratio test directly on the original sum; the limit to evaluate is a bit more complicated, but still very doable.

We need not apply the ratio test, but rather rely on the comparison test. All we need is to make use of the binomial theorem and elementary analysis. To that end we proceed.

First we take $$n>3$$. Then, from the binomial theorem we see that

\begin{align} 2^n&=\sum_{k=0}^n \binom{n}{k}\\\\ &\ge \binom{n}{4}\\\\ &=\frac1{24}n(n-1)(n-2)(n-3)\tag1 \end{align}

Using $$(1)$$, we can write for $$n\ge 4$$

\begin{align} \frac{n^2}{2^n+1}&\le \frac{n^2}{2^n}\\\\ &\le \frac{n^2}{\frac1{24} n(n-1)(n-2)(n-3)}\\\\ &\le \frac{32}{(n-2)(n-3)}\tag2 \end{align}

If we now restrict $$n$$ so that $$n\ge 6$$, then $$(n-2)(n-3)\ge \frac14 n^2$$. Using this estimate in $$(2)$$ reveals that

$$\frac{n^2}{2^n}\le \frac{128}{n^2}$$

Inasmuch as the series $$\sum_{n=6}^\infty \frac1{n^2}$$ converges, then by comparison the series of interest converges also. And we are done!

• Your answer is perfect, I just did not see it at the time and I apologize. Other than that, your answer was just what I was looking for. – Random Student Mar 25 at 18:01
• Pleased to hear. And feel free to up vote and accept an answer as you see fit. – Mark Viola Mar 25 at 18:02

IMO ratio test is straight forward: $$\frac{\frac{(n+1)^2}{2^{n+1}+1}}{\frac{n^2}{2^n+1}}= \frac{(n+1)^2}{n^2}\cdot \frac{2^n+1}{2^{n+1}+1} =\left(1+\frac{1}{n}\right)^2\cdot \frac{1+\frac{1}{2^n} }{2\cdot \left(1+\frac{1}{2^{n+1}} \right)}\stackrel{n \to \infty}{\longrightarrow}\frac{1}{2}$$

For each $$n\in\mathbb Z^+$$, $$\dfrac{n^2}{2^n+1}\leqslant\dfrac{n^2}{2^n}$$. On the other hand, if $$n$$ is large enough, then $$\dfrac{n^2}{2^n}\leqslant\left(\dfrac23\right)^n$$ because this means that $$n^2\leqslant\left(\dfrac43\right)^n$$ and an exponential function with a base greater than $$1$$ always growths faster than a polynomial function. So, since $$\displaystyle\sum_{n=0}^\infty\left(\dfrac23\right)^n$$ converges, your series converges too.

• How do you figure that $\frac{2}{3}^n$ is the best comparison with the series? – Random Student Mar 19 at 3:15
• I never claimed it is the best. Any number $q\in\left(\frac12,1\right)$ would have worked equally well. – José Carlos Santos Mar 19 at 7:16