# Check whether the following series is convergent/divergent

I am stuck on the following problem :

I have to check whether the following series is convergent/divergent?

$\,\, \frac{1}{1!}+\frac{3}{2!}+\frac{5}{3!}+....$

Take $u_n=\frac{2n-1}{n!}$ as the n-th term of the series, $a_n=\frac{n}{n!}$ so that $\frac{u_n}{a_n}=2-\frac1n \to 2$ as $\, n \to \infty.$

I was trying to use comparison test but I am unable to prove whether $\sum{a_n}$ converges /diverges. What I see that for $n >2, a_n=\frac{n}{n!}=\frac1{(n-1)!} > 1/(n-1)^{n-2}$. Now, I am not sure how to proceed further .

For what values of $x \, ,\sum\frac{a^n}{a^n+x^n}$ divergent ?

For $a >x, u_n^{1/n} >a. 1/(2a^n)^{1/n}=1/2^{\frac1n}$ and so $u_n^{1/n} \ge 1$ for infinitely many $n$ as $2^{\frac1n} \to 1$ as $n \to \infty.$

So, by Cauchy's test ,the series is divergent for $a>x.$

Can someone check it?

• Use integral test on the $1/(n-1)^{n-2}$. – Alex Jan 15 '16 at 16:06

## 2 Answers

For first problem, observe that $$(n-1)!=1\cdot 2\cdot 3 \ldots (n-1)>1\cdot 2\cdot 2 \ldots 2 =2^{n-2}$$ or $$\frac{1}{(n-1)!}<\frac{1}{2^{n-2}}$$ And then use the p-series test and you will find that $\frac{1}{(n-1)!}$ is convergent and so is $u_n$.

For the second problem, $$\sum_{n=0}^\infty \frac{a^n}{a^n+x^n}=\sum_{n=0}^\infty \frac{1}{1+(\frac{x}{a})^n}$$

Your solution is right, provided $a>1$.

• Thanks a lot sir for your time. But what about the second problem ? Is it ok ? – learner Jan 15 '16 at 16:22
• @learner Your second solution is correct, provided $a>1$. I have added that in the answer. – SchrodingersCat Jan 15 '16 at 16:28

Using ratio test, $$\lim_{n\to\infty}\dfrac{u_{n+1}}{u_n}=\lim_{n\to\infty}\dfrac{2n+1}{(2n-1)(n+1)}=?$$

In fact, $$\sum_{r=1}^\infty\dfrac{2r-1}{r!}=2\sum_{r=1}^\infty\dfrac1{(r-1)!}-\left(\sum_{r=0}^\infty\dfrac1{r!}-\dfrac1{0!}\right)$$

Now $\sum_{r=0}^\infty\dfrac1{r!}=e$