Finite sum of the series $\sum_{x =1}^{N} x e^{-\lambda x}$?

Question

Any idea to find the value of this finite series for $\lambda \geq 0$?

Answer 1

Using $$\sum_{k=0}^{n} t^{k} = \frac{1 - t^{n+1}}{1-t}$$ then $$\sum_{k=0}^{n} e^{- \lambda k} = \frac{1 - e^{-\lambda (n+1)}}{1-e^{- \lambda}}.$$ Taking a derivative with respect to $\lambda$ leads to $$\sum_{k=0}^{n} k \, e^{- \lambda \, k} = \frac{e^{-\lambda}}{(1-e^{-\lambda})^2} \, \left[ 1 - (n+1) \, e^{- \lambda \, n} + n \, e^{- \lambda \, (n+1)} \right].$$

Note that if $\lambda$ is allowed to equal zero the series becomes $$\sum_{k=0}^{n} k = \binom{n+1}{2}$$

Answer 2

$\textbf{hint}$ $$ \frac{d}{d\lambda}\sum_{x=0}^N\mathrm{e}^{-\lambda x} = \sum_{x=0}^N\frac{d}{d\lambda}\mathrm{e}^{-\lambda x} = -\sum_{x=0}^Nx\mathrm{e}^{-\lambda x} $$ We can use the left hand side to derive the one that we want. Also, note that we have a simple geometric series in the first equality.

Answer 3

Let $t=e^{-\lambda}$.

$$S_N:=t+2t^2+3t^3+\cdots Nt^N=t\left(1+2t+3t^2+\cdots Nt^{N-1}\right)\\ =t\left(t+2t^2+\cdots(N-1)t^{N-1}+1+t+t^2+\cdots t^{N-1}\right)$$

and

$$S_N=t\left(S_N-Nt^N+\frac{t^N-1}{t-1}\right).$$ You can draw $S_N$.