Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\frac{9}{4}.$

share|cite|improve this question
... did you mean numerator? – Nicolas Villanueva Jul 18 '11 at 12:45
Pretty nearly a duplicate of question 50919, which has 5 answers already, so have a look there. – Gerry Myerson Jul 18 '11 at 12:48
A related post:… I also add question mentioned by @Gerry Myerson, so that they are linked to each other: – Martin Sleziak Jul 18 '11 at 13:03
This is at the moment the only question tagged series Perhaps infinite-series would be better or maybe this tag could be omited completely. – Martin Sleziak Jul 18 '11 at 13:08
@Martin, good idea. I've deleted the "series" tag. I didn't add "infinite-series" since the question already has the "sequences-and-series" tag, which I think will suffice. – Gerry Myerson Jul 18 '11 at 23:37
up vote 10 down vote accepted

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 $$ \sum\limits_{k = 1}^\infty {\frac{{k + 1}}{{3^k }}} = \frac{5}{4}. $$

share|cite|improve this answer
This approach is of general interest, and can be used for confirmation. – Shai Covo Jul 18 '11 at 14:14
+1 For giving a nice solution to a slightly boring problem! ;) – AD. Jul 18 '11 at 19:36
@AD. Thanks. To make the solution more interesting, here is a probabilistic derivation of the formula ${\rm E}(X)=1/p$: $X=1$ with probability $p$, and $X = 1 + \tilde X$ with probability $1-p$, where $\tilde X$ is an independent copy of $X$. Hence, ${\rm E}(X) = p\cdot 1 + (1-p)(1 + {\rm E}(X))$. Solving for ${\rm E}(X)$ gives ${\rm E}(X) = 1/p$. – Shai Covo Jul 18 '11 at 20:08

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

share|cite|improve this answer

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.

share|cite|improve this answer
what does $+\cdots\infty$ mean? – wildildildlife Jul 18 '11 at 19:29
@wildildildlife: Well it just means that there are infinite number of terms in the given series. – user9413 Jul 18 '11 at 19:41
Perhaps, I should remove that – user9413 Jul 18 '11 at 19:42

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?

share|cite|improve this answer

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 $$\therefore \frac{2}{3}S=\frac{1}{2},S=\frac{3}{4}$$


share|cite|improve this answer

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 $f\bigl({1\over3}\bigr)={5\over4}$.

share|cite|improve this answer

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 $$\lim_{n\to\infty}F_n(x).$$

share|cite|improve this answer

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 $$\lim_{n\to\infty} \sum_{k=1}^n \frac{k+1}{3^k}=\frac{-\left(\frac{1}{3}\right)^2+2\cdot\frac{1}{3}}{\left(\frac{2}{3}\right)^2}=\frac{5}{4}$$

share|cite|improve this answer

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, $$ = \frac{3}{2} + \left(\frac{3}{2} - S_0 \right) + \left(\frac{3}{2} - S_1 \right) + \left(\frac{3}{2} - S_2 \right) \dots$$ $$ = \frac{3}{2} + \frac{1}{2} \left( 1 + \frac{1}{3} + \frac{1}{3^2} + \dots \right) $$ $$ = \frac{3}{2} + \frac{1}{2}S = \frac{3}{2} + \frac{1}{2} \frac{3}{2}$$ $$ = \mathbf{\frac{9}{4}}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.