An easy way to solve this limit of a sum?

Write out the infinite sum: $$S=\frac1{10^0}+\frac2{10^1}+\frac3{10^2}+\frac4{10^3}+\dots$$ Divide by ten and subtract from $S$: $$\frac1{10}S=\frac1{10^1}+\frac2{10^2}+\frac3{10^3}+\frac4{10^4}+\dots$$ $$S-\frac1{10}S=\frac1{10^0}+\frac1{10^1}+\frac1{10^2}+\frac1{10^3}+\dots$$ This is a geometric series, whose sum can be easily calculated: $$\frac9{10}S=\frac1{1-1/10}=\frac{10}9$$ $$S=\frac{100}{81}$$

When $x|<1$ $$\frac{1}{1-x}=\sum_{k=0}^{\infty} x^k ~~~~(1).$$ Differentiate (1) w.r.t. $x$ to get $$\frac{1}{(1-x)^2}= \sum_{k=0}^{\infty} k x^{k-1} \Rightarrow \frac{x.}{(1-x)^2}= \sum_{k=0}^{\infty} k x^{k}~~~~(2)$$ Let us put $x=1/10$ in (1) and (2) and then add them to get $$\sum_{k=0}^{\infty} \frac{k+1}{10^k}=\frac{10}{9}+\frac{10}{81}=\frac{100}{81}.$$