Proving the convergence of the series $\sum_{j=1}^\infty \frac{(2^j)^2}{j!}$ without root or ratio test

I am given the series: $\displaystyle \sum_{j=1}^\infty \frac{(2^j)^2}{j!}$ and I can show that it converges by using the ratio test, but I'm not sure how to approach to prove its convergence without it.

• Why do you want to prove convergence without the ratio test? – zhw. Aug 1 '16 at 2:49
• Solving through problems in a real analysis book and stuck on a question. – johnson Aug 1 '16 at 2:57

Let $a_j =\frac{(2^j)^2}{j!}$. Then $\frac{a_{j+1}}{a_j} =\frac{\frac{(2^{j+1})^2}{(j+1)!}}{\frac{(2^j)^2}{j!}} =\frac{2^{2(j+1)-2j}}{j+1} =\frac{4}{j+1}$.

Therefore, for $j \ge 9$, $\frac{a_{j+1}}{a_j} \le \frac12$.

By induction, for $j \ge 9$ and $k \ge 1$, $\frac{a_{j+k}}{a_j} \le \frac1{2^k}$.

Therefore for $j \ge 9$ and $k \ge 1$, $a_{j+k} \le a_j\frac1{2^k}$ so that, for $j \le 9$, $\sum_{k=1}^{\infty} a_{j+k} \le \sum_{k=1}^{\infty} a_j\frac1{2^k} =a_j$.

Since initial terms of a sum do not affect the convergence, the sum converges.

It can be done by explicit comparison with some nice series. For the sum from $8$ to $\infty$, note that the top is $4^j$ and the bottom is greater than $8^{j-7}$, that is, $8^{-7}8^j$. Thus for $j\ge 8$ we can compare with $8^7\sum_8^\infty \left(\frac{4}{8}\right)^j$.

Remark: This comparison is in principle not far from the Ratio Test, which fundamentally is also a comparison with a geometric series.

• And for $n\ge9$, $\dfrac{4^n}{n!}<\left(\frac{2}{5}\right)^{n-9}$ – John Wayland Bales Aug 1 '16 at 2:49
• @JohnWaylandBales: For no good reason, I decided to compare with the "canonical" geometric series $\sum \frac{1}{2^n}$. Lots of options. – André Nicolas Aug 1 '16 at 2:53
• I thought in order to compare, it had to be greater than all n, not just from 8 or 9? – johnson Aug 1 '16 at 2:57
• @AndréNicolas I noticed that $a_n$ was first less than $1$ when $n=9$ and that $\dfrac{a_{10}}{a_9}=\frac{2}{5}$ and the rest was trivial. – John Wayland Bales Aug 1 '16 at 3:01
• @MathematicianInWorks If it converges from $9$ onward, then it converges, since the sum of the first eight terms is finite. – John Wayland Bales Aug 1 '16 at 3:02

We have: $\dfrac{j^2}{j!}= \dfrac{j}{(j-1)!}= \dfrac{1}{(j-2)!}+ \dfrac{1}{(j-1)!}$, and you can reindex to show each of them converges, and they are easy to show to be convergent by comparing each with a $p>1$ -series.

If you have the series with $2^{2j} = 4^j$, then you would use $e^4$ as the targeting sum for the series. So either series is convergent hand down.

• hey terribly sorry, I made a typo in my initial question. It was supposed to be 2^j, not 2*j – johnson Aug 1 '16 at 2:11

If you know $\sum 1/j^2 < \infty,$ you can argue like this: For $j\ge 6,$

$$\frac{4^j}{j!} = \frac{4}{j}\frac{4}{j-1} \cdots \frac{4}{4}\frac{4}{3}\frac{4}{2}\frac{4}{1}\le \frac {4^6}{4!}\frac{1}{j(j-1)}.$$

Dividing the term on the right by $1/j^2$ and letting $j\to \infty$ gives a finite limit. Thus the limit comparison test shows our series converges.

• @zhw...any reason to divide it by $1/j^2$ ? – Mathlover Aug 1 '16 at 16:11
• Yes, I am assuming as known $\sum 1/j^2< \infty.$ – zhw. Aug 1 '16 at 16:30
• $\sum_{1}^{\infty}\frac{4^j}{j!}=$finite sum$+\sum_{6}^{\infty}\frac{4^j}{j!}$. Now the summation on RHS is less than or equal to $\sum_{6}^{\infty}\frac{4^6}{4!}\frac{1}{j(j-1)}$$\approx\sum_{6}^{\infty} \frac{1}{j^2}$. – Mathlover Aug 1 '16 at 16:51
• And you know that because by dividing the terms of the first series by those of the second you have a sequence that has a finite positive limit. – zhw. Aug 1 '16 at 18:47
• @zhw...Is my comment not sufficient to prove the result? Where is the fallacy? – Mathlover Aug 2 '16 at 1:19