I would like to approximate the divergent sum $$ \sum_{i=0}^n \frac{e^{\mu i}}{1+i}\ , $$ where $\mu >0$. I attempted to use integration by parts, arrived at $$ \int_0^n \frac{e^{\mu t}}{1+t} dt = e^{\mu n} \log{(1+n)} - \frac{1}{\mu}\int_0^n e^{\mu t}\log{(1 + t)} dt\ , $$ and promptly got stuck. Is there a way to proceed? Or perhaps a better approach?

Context for the curious: this sum would be the capital recovery factor in an investment appraisal where the return has a trend $\mu$, but where standard exponential discounting has been substituted with hyperbolic discounting.

  • 1
    $\begingroup$ In the summation, is 'n' the final value, or the summation variable? $\endgroup$ Oct 20, 2020 at 14:52
  • 2
    $\begingroup$ Do you mean $$\sum _{i=0}^n \frac{e^{i \mu }}{i+1}$$ $\endgroup$
    – Raffaele
    Oct 20, 2020 at 14:58
  • 1
    $\begingroup$ The variable must be t. I do not see any t in the fraction or in upper bound of interval. $\endgroup$
    – sirous
    Oct 20, 2020 at 15:06
  • $\begingroup$ Thank you for the comments, I've fixed the question. $\endgroup$
    – Anthony
    Oct 20, 2020 at 17:18

1 Answer 1


You could do a change of variables $t \to t/\mu -1$. Then, the Integral takes the for of the exponential integral function with a pre-factor and different argument: $$ \int_0^n \frac{e^{\mu t}}{1 + t} dt = e^{-\mu} \int_1^{\mu(n+1)} e^t/t~ dt = e^{-\mu} \Big( \text{Ei}[\mu(n+1)] - \text{Ei}[1] \Big) $$

Does that help?

Alternatively: We know that the exponential function is increasing very rapidly. So rapidly, in fact, that the integral over all values up to $x$ is about as large as $e^x$ itself. Maybe this holds for the sum as well: $$\int_0^x e^{\mu t} dt \approx \mu^{-1} e^{\mu x} \overset{?}{\rightarrow} \sum_{i=0}^n e^{i \mu} \approx \mu^{-1} e^{\mu (n+1)}.$$ If so, then the $1/(1+i)$ should not change much: $$\sum_{i=0}^n \frac{e^{\mu i}}{i+1} \approx \mu^{-1} \frac{e^{\mu (n+1)}}{n+2}. $$ I tried a few examples and it seems to give at least the right order of magnitude for $n\mu$ up to 100. That means about $50 \%$ error, but such deviations already appears when comparing $\int e^t dt$ and $\sum_i e^i$. So a better approximation can probably not be achieved through the integral but only by looking at the sum directly.

  • $\begingroup$ Thanks for this. This is indeed helpful since the Ei function appears to have several approximations. $\endgroup$
    – Anthony
    Oct 22, 2020 at 8:43
  • $\begingroup$ That was my hope, but I have to admit that it is not easy to find one that does not rely on numerical results. If you find a good one, let me know! But maybe I have an intuitive solution to this... I'll edit my answer. $\endgroup$
    – Cream
    Oct 22, 2020 at 14:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .