Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the value of :

$$\sum_{n=1}^{\infty}\frac{n^2+n+1}{3^n}$$

share|improve this question
1  
I don't understand what's with this fashion in downvoting people. Is it a good thing to downvote a new person? What's the reason in that? (+1) –  Chris's sis Sep 22 '12 at 11:21
    
It would be helpful if you provided some reason for asking this, and your attempts so far. If it is homework, you should tag it as such. –  Feanor Sep 22 '12 at 11:22
    
@Chris'ssister: My guess is that it's because this kind of question has been asked many times before on this site, and the answer could have been found by a bit of searching. –  Hans Lundmark Sep 22 '12 at 20:47
    
For example here: math.stackexchange.com/questions/30732/… –  Hans Lundmark Sep 22 '12 at 20:48

4 Answers 4

up vote 5 down vote accepted

You have for $|x|<1$

$$ \sum_{k=0}^\infty x^k=\frac{1}{1-x}$$

$$ \sum_{k=0}^\infty kx^k=x\cdot \left(\frac{1}{1-x}\right )'$$

$$ \sum_{k=0}^\infty k^2x^k=x \cdot \left( x \cdot \left( \frac{1}{1-x}\right )'\right )'$$

Replace $x$ with $1/3$ and you will get the result.

share|improve this answer

HINT: you may use $e^{kx}$. Then solve the geometrical progression, derive once/twice its both sides and then plug in $x=-\ln(3)$.

share|improve this answer

I don't know the exact steps of how to get that, but i figured out that this comes out to be $\frac 11+\frac 79+\frac {15}{27}+\ldots$

and this link here, gives the solution to be $\frac {11}4$. Refer WolframAlpha Solution.

share|improve this answer

In most practical applications, you can just ask Mathematica, and it will tell you it's $\frac{11}{4}$.

If you want to arrive at the formula in a more rigorous way, you can do the following:

Consider the function $f$: $$f(x) = \frac{1}{1 - x} = \sum_{n=0}^\infty x^n$$

An initial observation is that $f(\frac{1}{3}) = \sum_{n=0}^\infty \frac{1}{3^n}$ so it looks a little like your sum. Now, consider $f'$: $$f'(x) = \frac{1}{(1 - x)^2} = \sum_{n=1}^\infty n x^{n-1} = \sum_{n=0}^\infty (n+1) x^{n} $$

When you plug in $\frac{1}{3}$ again, you find $f'(\frac{1}{3}) = \sum_{n=0}^\infty \frac{n+1}{3^n}$. Finally, consider $f''$: $$f''(x) = \frac{2}{(1 - x)^3} = \sum_{n=0}^\infty (n+2)(n+1) x^{n} = \sum_{n=0}^\infty (n^2 + 3n + 2) x^{n}$$ When you plug in $\frac{1}{3}$ once more, you find $f''(\frac{1}{3}) = \sum_{n=0}^\infty \frac{n^2 + 3n + 2}{3^n}$. Now, all you have to do is to express $n^2 + n +1$ as: $$ n^2 + n + 1 = (n^2 + 3n + 1) - 2 (n+1) + 1 $$ so you get that the sought sum $S$ is: $$ S = f''\left(\frac{1}{3}\right) - 2 f'\left(\frac{1}{3}\right) - f\left(\frac{1}{3}\right)$$ The remaining computation is not pleasant, but it is definitely doable, and does not involve any new creative ideas.


I was a little too careless: the computation I did was for a sum raging from $n=0$ rather than $n=1$ as in the problem. It is however easy to mend - just subtract the initial term, which is just $\frac{1}{3}$.

share|improve this answer

Your Answer

 
discard

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.