Let $\displaystyle f(x)=\lim_{n\to\infty} \left(\frac{ax}{n}\left(\sum_{k=1}^n \frac{[k^2-e^{-x}+k-1]}{k(k+1)}\right)\right)+\lambda$

Find $f(x)$ if $[\,.]$ denotes G.I.F.

I know how to solve such type of questions (by splitting the integral at integers), but that, too, in case where argument of G.I.F. is quite simple. I've never encountered any problem as mentioned above. Here's the original question.

I just want to know how to tackle these type of questions.

Any hints are appreciated!

  • 3
    $\begingroup$ I think it is probably easiest to break apart the function based on $x$. For example, for $x>0$ we have $0<e^{-x}<1$ so that $$[k^2-e^{-x}+k-1]=k^2+k-2.$$ I haven’t worked out any details, but this seems like a promising start. $\endgroup$
    – Clayton
    Jun 18, 2020 at 12:46

1 Answer 1


By the hint given by @Clayton, I've solved the problem myself and that was truly a promising start.

\begin{align} f(x)&=\lim_{n\to\infty} \frac{ax}{n}\left(\sum_{k=1}^n \frac{[k^2-e^{-x}+k-1]}{k(k+1)}\right)+\lambda\\ &=\lim_{n\to\infty} \frac{ax}{n}\left(\sum_{k=1}^n \frac{k^2+k-2}{k(k+1)}\right)+\lambda\\ &=\lim_{n\to\infty} \frac{ax}{n}\left(\sum_{k=1}^n 1-2\left(\frac{1}{k}-\frac{1}{k+1}\right)\right)+\lambda\\ &=\lim_{n\to\infty} \frac{ax}{n}\left(n+\frac{1}{n+1}-2\right)+\lambda\\ &=ax+\lambda. \end{align}


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