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Let $$f(n,k) = \dfrac{1! + 2! + ... + n!}{(n+k)!}$$

Find (if exists) the values of $k$ such that $\sum_{n=1}^{\infty} f(n,k)$ converges.


What I've done so far:

  1. If exists $k$ such that $\sum_{n=1}^{\infty} f(n,k)$ converges, then $\sum_{n=1}^{\infty} f(n,M)$ converges for all $M\ge k$ That is because $f(n,k+1) = \dfrac{1}{n+k+1} f(n,k)$ and apply Abel's Test.

  2. For $k=1$ the series is divergent:

In general, I found that $$f(n+1,k) = \dfrac{1}{n+k+1} f(n,k) + \dfrac{(n+1)!}{(n+k+1)!}$$ Then, if $k=1$, $$f(n+1,1) = \dfrac{1}{n+2}f(n,1) + \dfrac{1}{n+2}$$

Then comparing with $\dfrac{1}{n+2}$ we have that the series diverges.

I think that for $k=2$ the series is convergent, but I'm missing something...

Thanks for the help!

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By the inequalities

$$n!\le 1!+2!+\cdots+n!\le (n-2)(n-2)!+(n-1)!+n!$$ and using the squeeze theorem we see easily that

$$1!+2!+\cdots+n!\sim_\infty n!$$ hence

$$f(n,k)\sim_\infty \frac{1}{(n+k)\cdots(n+1)}$$ hence

  • if $k=1$ then $f(n,1)\sim\frac1{n+1}$ and then the series is divergent
  • if $k\ge2$ then $f(n,k)=O\left(\frac1{n^2}\right)$ and then the series is convergent.
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For $k = 2$ $$f(n,k) \leq (n-1)\dfrac{(n-1)!}{(n+2)!} + \dfrac{n!}{(n+2)!} = \dfrac{n-1}{n(n+1)(n+2)} + \dfrac{1}{(n+1)(n+2)}$$

Then since $\sum \dfrac{1}{n^2}$ converges, we can easily go from here....

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