Analyze if this series converges: $\sum_{n=0}^{\infty}\frac{n^{2}+1}{n!}$

Question

Analyze if this series converges: $\sum_{n=0}^{\infty}\frac{n^{2}+1}{n!}$

I have used ratio test: $\lim_{n\rightarrow \infty}\left |\frac{a_{n+1}}{a_{n}} \right |< 1$

$\Rightarrow$

$\lim_{n\rightarrow \infty}\left | \frac{(n+1)^{2}+1}{(n+1)!}:\frac{n^{2}+1}{n!} \right | = \lim_{n\rightarrow \infty}\frac{((n+1)^{2}+1)n!}{(n+1)!*(n^{2}+1)}$

$= \lim_{n\rightarrow \infty}\frac{(n+1)^{2}+1}{(n+1)*(n^{2}+1)} = \lim_{n\rightarrow \infty}\frac{n^{2}+2n+2}{n^{3}+n^{2}+n+1}$

The denominator is bigger than the enumerator, so we got $\infty > 1$ and thus the sequence will diverge. (Can I just say that or this requires an additional proof? We got a $n^{2}$ in the enumerator and a $n^{3}$ in the denominator...?)

Did I do it correctly?

Edit: Converges absolutely to $0$ and NOT $\infty$

Answer 1

Just to provide a concrete underpinning to this general reasoning: $$\sum_{n=0}^{\infty}\frac{1}{n!}x^n=e^x$$ Differentiate and then multiply by $x$: $$\sum_{n=0}^{\infty}\frac{n}{n!}x^n=xe^x$$ Differentiate and then multiply by $x$: $$\sum_{n=0}^{\infty}\frac{n^2}{n!}x^n=x(x+1)e^x$$ Adding the first and last equations: $$\sum_{n=0}^{\infty}\frac{n^2+1}{n!}x^n=(x^2+x+1)e^x$$ So the value of your sum is $3e$.

Answer 2

You made a mistake at the final step to make a conclusion. :)

Since $\lim_{n\rightarrow \infty}\left |\frac{a_{n+1}}{a_{n}} \right |=0$ then the series converges absolutely.

Also, we don't know the value of the sum so it is still wrong to say it converges absolutely to $0$! The ratio test just tells us that the limit

$$\lim_{m \to \infty} \sum_{n=0}^{m}\frac{n^{2}+1}{n!}$$

exists but it says nothing about the value of the limit! :)

Answer 3

HINT Observe that $$ \lim_{n\to\infty}\frac{n^{2}+2n+2}{n^{3}+n^{2}+n+1}=\lim_{n\to\infty}\frac{\frac{1}{n}+\frac{2}{n^2}+\frac{2}{n^3} }{ 1+\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}}=\frac{0}{1}=0 $$

Answer 4

The limit of a rational function (quotient of two polynomials) at $\infty$ is the limit of the ratio of the highest degree terms, namely $$\lim_{n\to\infty}\frac{n^{2}+2n+2}{n^{3}+n^{2}+n+1}=\lim_{n\to\infty}\frac{n^{2}}{n^{3}}=\lim_{n\to\infty}\frac{1}{n}=0.$$

Answer 5

The main term goes to zero so fast that any crude comparison gives absolute convergence.
Moreover, the value of the series is simple to compute: $$ \sum_{n\geq 0}\frac{n^2+1}{n!}=\sum_{n\geq 0}\frac{1}{n!}+\sum_{n\geq 0}\frac{n(n-1)+n}{n!}=\sum_{n\geq 0}\frac{1}{n!}+\sum_{n\geq 1}\frac{1}{(n-1)!}+\sum_{n\geq 2}\frac{1}{(n-2)!} = \color{red}{3e}.$$