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Given a series, how does one calculate that limit below? I noticed the numerator is an arithmetic progression and the denominator is a geometric progression — if that's of any relevance —, but I still don't know how to solve it.
$$\lim_{n\to\infty} \sum^n_{k=0} \frac{k+1}{3^k}$$
I did it "by hand" and the result should be $\frac{9}{4}.$
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.
By the way, the answer you got by hand is off a bit. Hopefully my hint will help you derive the correct answer.
First, note that by definition, $$\sum_{k=1}^\infty a_k=\lim_{n\rightarrow\infty}\;\sum_{k=1}^na_k$$ so I will use the infinite sum as a shorthand.
We know that $$f(x)=\sum_{k=1}^\infty\frac{1}{x^{k+1}}=(x^{-1})^2+(x^{-1})^3+\cdots=\frac{{x}^{-2}}{1-x^{-1}}=\frac{1}{x^2-x}.$$ What does that mean $$g(x)=-x^2f'(x)$$ is? Assume (or, better, prove) that differentiation can split up over an infinite sum. Can you use this to help your computations?
divide that formular like this.
$$\sum_{k=1}^{\infty}\left (\frac{k}{3^k}+\frac{1}{3^k}\right )$$
then $$\sum_{k=1}^{\infty}\frac{k}{3^k}+\frac{\frac{1}{3}}{1-\frac{1}{3}}=\sum_{k=1}^{\infty}\frac{k}{3^k}+\frac{1}{2}$$ power series $$ \sum_{k=1}^{\infty}\frac{k}{3^k}$$
let $$ \sum_{k=1}^{\infty}\frac{k}{3^k}=S$$ Then,
$$ S=1\times \frac{1}{3}+2\times \frac{1}{3^2}+3\times \frac{1}{3^3}+\cdots \cdots $$ $$\frac{1}{3}S=1\times \frac{1}{3^2}+2\times \frac{1}{3^3}+3\times \frac{1}{3^4}+\cdots \cdots $$ $$S-\frac{1}{3}S=1\times \frac{1}{3}+(2-1)\times \frac{1}{3}+(3-2)\times \frac{1}{3}\cdots \cdots =\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}$$ Answer is $$\therefore \frac{2}{3}S=\frac{1}{2},S=\frac{3}{4}$$
$$\frac{3}{4}+\frac{1}{2}=\frac{5}{4}$$
If you are given a "constant series" $\sum_{k=1}^\infty a_k$ of this kind (maybe even with factors $k!$ in the denominator) it is often helpful to introduce a factor $x^k$ into the general term. We then are speaking of a function $f(x):= \sum_{k=1}^\infty a_k x^k$ and want to know the value $f(1)$. Note that now we have new tools at our disposal, namely differentiation or integration with respect to $x$, multiplication by $x$ or ${1\over x}$, replacing $x^2$ by $u$, etc. By means of such operations it is then often possible to transform the power series $\sum_{k=1}^\infty a_k x^k$ into a series that we recognize as the series of a familiar function like ${1\over 1-x}$, $\cosh x$, etc.
In the example at hand we can subsume the factors ${1\over 3^k}$ into the $x^k$ and compute the value $f\bigl({1\over3}\bigr)$ at the end. This means we are now considering the function $$f(x):=\sum_{k=1}^\infty (k+1) x^k\ .$$ Looking at this formula we see that $f(x)=g'(x)$ for the function $$g(x):=\sum_{k=1}^\infty x^{k+1}=x^2+x^3+x^4+\ldots = {x^2\over 1 -x}\ ,$$ and this is valid for all $x$ of absolute value $<1$. It follows that $$f(x)=g'(x)={2x -x^2\over (1-x)^2}\ ;$$ therefore the value we want is $f\bigl({1\over3}\bigr)={5\over4}$.
The following is a variant presentation of a standard solution. For now we omit convergence considerations. We include the term that has $x$ raised to the $0$-th power, because it wants to be included. Let $$F(x)=1+2x+3x^2+4x^3+ \cdots +nx^{n-1}+\cdots.$$ Multiply $F(x)$ by $(1-x)$. So $$(1-x)F(x)=(1-x)(1+2x+3x^2+4x^3+ \cdots +nx^{n-1}+\cdots).$$ Multiplying out the right-hand side takes some concentration. I think it is called long multiplication.
But we quickly notice that the product is $1+x+x^2+\cdots +x^n+\cdots$, and conclude that $$(1-x)F(x)=\frac{1}{1-x}.$$
The finite sum case: Let $$F_n(x)=1+2x+3x^2+ \cdots +nx^{n-1}.$$ Multiply both sides by $1-x$. We obtain $$(1-x)F_n(x)= (1-x)(1+2x+3x^2+ \cdots +nx^{n-1})=1+x+x^2+\cdots+x^{n-1}-nx^n.$$ Thus, if $x \ne 1$, then $$(1-x)F_n(x)=\frac{1-x^n}{1-x}-nx^n.$$
If we wish, when $|x|\lt 1$, we can now compute $$\lim_{n\to\infty}F_n(x).$$
Firstly, what you have is a limit of a finite sum, and the limit is a series:
$$\lim_{n\to\infty} \sum_{k=1}^n \frac{k+1}{3^k}:=\sum_{k=1}^\infty \frac{k+1}{3^k}$$
Now, to the question you asked: there are a few tricks you can use:
1) You know that $\sum_{k=1}^nx^k=\frac{1-x^{n+1}}{1-x}-1$ (geometric progression).
2) That implies, by deriving both sides, that $\sum_{k=1}^nkx^{k-1}=\frac{-(n+1)x^{n}(1-x)+(1-x^{n+1})}{(1-x)^2}=\frac{nx^{n+1}-(n+1)x^{n}+1}{(1-x)^2}$
3) From here you get that $\sum_{k=1}^n(k+1)x^k=\frac{1-x^{n+1}}{1-x}-1+x\frac{nx^{n+1}-(n+1)x^{n}+1}{(1-x)^2}=\frac{(n+1)x^{n+2}-(n+2)x^{n+1}-x^2+2x}{(1-x)^2}$.
4) Now plug in $x=\frac{1}{3}$. As $n\to\infty$, you have $x^n\to 0$. This implies that $$\lim_{n\to\infty} \sum_{k=1}^n \frac{k+1}{3^k}=\frac{-\left(\frac{1}{3}\right)^2+2\cdot\frac{1}{3}}{\left(\frac{2}{3}\right)^2}=\frac{5}{4}$$
Expanding your problem:
$$ \sum\limits_{k = 0}^\infty {\frac{{k + 1}}{{3^k }}} = \frac{1}{3^0} + \frac{2}{3^1} + \frac{3}{3^2} + \frac{4}{3^3} + \dots $$
$$ = 1 + \left (\frac{1}{3} + \frac{1}{3} \right) + \left(\frac{1}{3^2} + \frac{1}{3^2} + \frac{1}{3^2} \right) + \left(\frac{1}{3^3} + \frac{1}{3^3}+ \frac{1}{3^3}+ \frac{1}{3^3}\right) + \dots$$
This can be grouped into:
$$ = \left(1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots\right)+ $$ $$ \left(\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \dots\right)+ $$ $$ \left(\frac{1}{3^2} + \frac{1}{3^3} + \dots\right)+ $$ $$ \left(\frac{1}{3^3} + \dots\right) + \dots $$
Using the fact that $ S = \sum_{n=0}^{\infty} \frac{1}{3^n} = \frac{3}{2}$: $$ = \frac{3}{2} + $$ $$ \frac{3}{2} - (1) + $$ $$ \frac{3}{2} - \left(1 + \frac{1}{3} \right) + $$ $$ \frac{3}{2} - \left( 1 + \frac{1}{3} + \frac{1}{3^2} \right ) + \dots $$
The partial sum $S_k$ is computed as: $S_k = \sum_{n=0}^k \frac{1}{3^n} = \frac{3}{2} - \frac{1}{2}\left(\frac{1}{3}\right)^k$
Hence, $$ = \frac{3}{2} + \left(\frac{3}{2} - S_0 \right) + \left(\frac{3}{2} - S_1 \right) + \left(\frac{3}{2} - S_2 \right) \dots$$ $$ = \frac{3}{2} + \frac{1}{2} \left( 1 + \frac{1}{3} + \frac{1}{3^2} + \dots \right) $$ $$ = \frac{3}{2} + \frac{1}{2}S = \frac{3}{2} + \frac{1}{2} \frac{3}{2}$$ $$ = \mathbf{\frac{9}{4}}$$
