# Approximate a Sum using Integration by Parts

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.

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

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: $$\int_0^n \frac{e^{\mu t}}{1 + t} dt = e^{-\mu} \int_1^{\mu(n+1)} e^t/t~ dt = e^{-\mu} \Big( \text{Ei}[\mu(n+1)] - \text{Ei}[1] \Big)$$
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.