Find $\Sigma_{1}^n ka_k$

Question

I was given $$a_n = \int _{0}^{1/4} x^n.e^{-x} dx (n=1,2,3..)$$ $$na_n = a_{n+1} - a_n + {\left( \frac 14 \right)}^{n+1} e^{-1/4}$$

I was asked, Find $\displaystyle\sum_{1}^n ka_k$

$$\sum_{1}^n ka_k = a_1 + 2a_2 + 3a_3 ... + na_n$$ $$na_n = a_{n+1} - a_n + {\left( \frac 14 \right)}^{n+1} e^{-1/4} = (n+1)a_n + \frac 54 e^{-1/4} - 1$$

What should i do now?

Answer 1

We have $$na_n=a_{n+1}-a_n+\frac{1}{4^{n+1}}e^{-1/4}$$ thus $$\sum_{k=1}^n ka_k =\sum_{k=1}^n a_{k+1}-a_k+\frac{1}{4^{k+1}}e^{-1/4}\\ =\sum_{k=1}^n a_{k+1}-a_k +\sum_{k=1}^n \frac{1}{4^{k+1}}e^{-1/4}\\ =(a_{n+1}-a_1)+e^{-1/4}\sum_{k=1}^n \frac{1}{4^{k+1}}\\ $$ The last sum is sum of geometric progression . Note that the first sum is telescopic.

Answer 2

Even without proving the recurrence relation, the series $\sum_{k\geq 1}ka_k$ can be computed via

$$\sum_{k\geq 1}ka_k = \int_{0}^{1/4} e^{-x}\sum_{k\geq 1}k x^k\,dx =\int_{0}^{1/4}\frac{x e^{-x}}{(1-x)^2}\,dx=\left[\frac{e^{-x}}{1-x}\right]_{0}^{1/4}=\color{red}{\frac{4}{3}e^{-1/4}-1}.$$

A similar argument applies to the partial sums of the LHS.