Limit of this series: $\lim_{n\to\infty} \sum^n_{k=0} \frac{k+1}{3^k}$?

Let $X$ be a geometric random variable with probability of success $p=2/3$, so that $$ {\rm P}(X=k)=(1-p)^{k-1}p = \frac{2}{{3^k }}, \;\; k=1,2,3,\ldots. $$ From the easy-to-remember fact that ${\rm E}(X)=1/p$, it follows that $$ \frac{3}{2} + 1 = {\rm E}(X) + 1 = {\rm E}(X + 1) = \sum\limits_{k = 1}^\infty {(k + 1){\rm P}(X = k) = 2\sum\limits_{k = 1}^\infty {\frac{{k + 1}}{{3^k }}} } . $$ Hence $$ \sum\limits_{k = 1}^\infty {\frac{{k + 1}}{{3^k }}} = \frac{5}{4}. $$

if you take the derivative of $$ \frac{1}{1-x}=\sum_{k=0}^{\infty}x^k $$ you get $$ \frac{1}{(1-x)^2}=\sum_{k=1}^{\infty}kx^{k-1} $$ evaluating at $x=1/3$ gives $$ \frac{1}{(1-1/3)^2}=\sum_{k=1}^{\infty}\frac{k}{3^{k-1}} $$ subtract off the $k=1$ term to get your series $$ 9/4-1=5/4 $$ so if you want the anwer to be $9/4$, you might want to change your $1$ to a $0$ in the indexing

For summing $\sum\frac{k}{3^k}$ use the following formula: $$(1-x)^{-2} = 1 + 2x + 3x^{2} + 4x^{3} + \cdots \qquad \Bigl[\because \small (1-x)^{-n} = 1+nx +\frac{n\cdot (n-1)}{2!}\cdot x^{2} + \cdots \Bigr]$$ Multiplying the above equation by $x$ and then putting $x=\frac{1}{3}$ we have $$\frac{1}{3} + \frac{2}{9}+\frac{3}{27} + \frac{4}{3^{4}} + \cdots = \frac{1}{3}\Bigl(1-\frac{1}{3}\Bigr)^{-2} = \frac{3}{4} \qquad\quad \cdots (1)$$

Also you know that $$\sum\limits_{k=1}^{\infty} \frac{1}{3^k}= \frac{\frac{1}{3}}{1-\frac{1}{3}} =\frac{1}{2} \qquad\qquad \cdots (2)$$

Add equations $(1)$ and $(2)$ to get your answer.