the series in the problem is as follows, and we would like to see if the series converges, and what is the sum of infinite series:

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

It's a homework problem, but I already asked my teacher about it at school and he didn't really help me that much.

The earlier problems that we had were easier than this one, because I cannot find the common ratio for this infinite series (infinite geometric series should have common ratio, and when you know it you can tell if there is convergence)

I read online after class, that what I should do is to find the limit of the sequence of partial sums. This limit, if it exists and is a real-number, should be the value of the sum of the infinite series.

It makes sense conceptually when you have a large value of k, for a finite series $S_{k}$, when k becomes larger the finite sum becomes more and more like the sum of infinite series.

So that if I have partial sums in a sequence $S_{1}, S_{2}, S_{3}...S_{k} $ The larger the k, the better it is representing infinite series, assuming it converges at all, towards anything.

I managed to put some large values of n, into wolfram alpha and the correct answer seems to be that the infinite series converges and sum of infinite series is 4.5

But I don't have a solid idea how I could find the explicit formula for the sequence of partial sums for this particular infinite series. And I don't particularly have any other good ideas how to prove that the sum of infinite series should be 4.5 in this case...


$$\sum_{n=1}^\infty\frac{2^{n+1}+1}{3^n}=2\sum_{n=1}^\infty\left(\frac23\right)^n+\sum_{n=1}^\infty\left(\frac13\right)^n.$$Can you take it from here?

  • $\begingroup$ Can you clarify where you get the idea to split the series like that, I'm not very good at the calculation rules for sums (sigma)... Indeed I only really studied about infinite series last week. I mean does there exist some algebraic formulas for manipulation of sums like that $\endgroup$ – Late347 Jan 25 '18 at 15:55
  • $\begingroup$ @Late347 Yes: if both series $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ converge, then the series $\sum_{n=1}^\infty(a_n+b_n)$ converges too and its sum is the sum of the two original series. $\endgroup$ – José Carlos Santos Jan 25 '18 at 15:57

After 20 years, I still have to rederive the identity $$ \sum_{n=1}^{\infty} \alpha^n = \frac{\alpha}{1-\alpha} \tag{1}$$ whenever I need it. Since it isn't that difficult, perhaps we should do that first.

Suppose that $|\alpha| < 1$, and let $$ S_k := \sum_{n=1}^{k} \alpha^k = \alpha + \alpha^2 + \dotsb + \alpha^{k} $$ be the $k$-th partial sum. Then $$ \alpha S_k = \sum_{n=1}^{k} \alpha^{k+1} = \alpha^2 + \alpha^3 + \dotsb + \alpha^{k+1}. $$ Subtracting, we obtain \begin{align} (1-\alpha) S_k &= S_k - \alpha S_k \\ &= \left( \alpha + \alpha^2 + \dotsb + \alpha^{k} \right) - \left( \alpha^2 + \alpha^3 + \dotsb + \alpha^{k+1} \right) \\ &= \alpha + \left( \alpha^2 - \alpha^2\right) + \left( \alpha^3 - \alpha^3\right) + \dotsb + \left(\alpha^k - \alpha^k\right) - \alpha^{k+1} \\ &= \alpha - \alpha^{k+1}. \end{align} Solving for $S_k$, we get $$ S_k = \frac{\alpha - \alpha^{k+1}}{1-\alpha}. $$ Since $|\alpha| < 1$, it follows that $\lim_{k\to\infty} \alpha^{k+1} = 0$, and so $$ \sum_{n=1}^{\infty} \alpha^n := \lim_{k\to \infty} \sum_{n=1}^{k} \alpha^n = \lim_{k\to\infty} \frac{\alpha - \alpha^{k+1}}{1-\alpha} = \frac{\alpha}{1-\alpha}. $$ But this is exactly the identity at (1). Huzzah!

Now, how do we use this here?

Observe that $$ \frac{2^{n+1} + 1}{3^n} = \frac{2^{n+1}}{3^n} + \frac{1}{3^n} = 2 \left( \frac{2}{3} \right)^n + \left( \frac{1}{3} \right)^n. \tag{2}$$ But series are linear, and so play nice with addition and multiplication. Specifically, if $\sum a_n$ and $\sum b_n$ converge, and $C$ is any constant, then we have $$ \sum_{n=1}^{\infty} \left[ a_n + b_n \right] = \sum_{n=1}^{\infty} a_n + \sum_{n=1}^{\infty} b_n \qquad\text{and}\qquad \sum_{n=1}^{\infty} Ca_n = C \sum_{n=1}^{\infty} a_n. $$ These are basic rules for "simplifying" series, which are going to be quite useful for you. If you are familiar with integration, you can perform exactly these manipulations if you replace the $\sum$ with $\int$, and the sequences with functions. If you are not familiar with integration, ignore this comment for now, but maybe keep it in the back of your mind for later. \begin{align} \sum_{n=1}^{\infty} \frac{2^{n+1} + 1}{3^n} &= \sum_{n=1}^{\infty} \left[ 2 \left( \frac{2}{3} \right)^n + \left( \frac{1}{3} \right)^n \right] && (\text{by application of (2)}) \\ &= 2\sum_{n=1}^{\infty} \left( \frac{2}{3} \right)^n + \sum_{n=1}^{\infty} \left( \frac{1}{3} \right)^n && (\text{simplify the series}) \\ &= 2 \cdot \frac{\frac{2}{3}}{1-\frac{2}{3}} + \frac{\frac{1}{3}}{1- \frac{1}{3}} && (\text{by application of (1)}) \\ &= 2 \cdot \frac{\frac{2}{3}}{1-\frac{2}{3}}\cdot \frac{3}{3} + \frac{\frac{1}{3}}{1- \frac{1}{3}}\cdot \frac{3}{3} && (\text{silly, pedantic arithmetic}) \\ &= 2 \cdot \frac{2}{3-2} + \frac{1}{3-1} && (\text{more pedantic arithmetic}) \\ &= \frac{4}{1} + \frac{1}{2} && (\text{and some more}) \\ &= \frac{8+1}{2} && (\text{almost there}) \\ &= \frac{9}{2}. && (\text{done!}) \end{align}

  • $\begingroup$ IT looks very understandable now, I jotted down my own version on a piece of paper here at my desk. My worries and further question is more about the justification of splitting the Series into two separate series... Doesn't there have to be the assumption such that the combined original series has to be convergent to begin with before manipulating? $ \sum_{n=1}^{\infty} \frac{2^{n+1}+1}{3^n}$ $\endgroup$ – Late347 Jan 25 '18 at 16:31
  • $\begingroup$ EDIT I just wanted to ask for sure, because I don't want to do silly mistakes with the infinite series... I still have a bad taste in my own mouth after having watched the numberphile video about summing divergent infinite series into each other (the video can be found on youtube with "1+2+3... = -1/12") $\endgroup$ – Late347 Jan 25 '18 at 16:33
  • $\begingroup$ The justification kind of comes in through the back door. If $\sum a_n$ and $\sum b_n$ are convergent, then $\sum (a_n + b_n)$ converges, and $\sum (a_n + b_n) = \sum a_n + \sum b_n$. In the argument above, we break up a series of the form $\sum (a_n + b_n)$ into two different convergent series $\sum a_n$ and $\sum b_n$, then use the fact that both of these converge in order to show that the original series converges. $\endgroup$ – Xander Henderson Jan 25 '18 at 16:39
  • $\begingroup$ On the other hand, if all you need to know is that the original series converges (and this might assuage your conscience regarding this assumption), you might note that $2^{n+1} + 1 < 2^{n+2}$ for all $n$, and $\sum \frac{2^{n+2}}{3^n}$ is a convergent geometric series. But then $$ 0 < \sum \frac{2^{n+1}+1}{3^n} \le \sum \frac{2^{n+2}}{3^n} < \infty, $$ which demonstrates that the original series does, in fact, converge. $\endgroup$ – Xander Henderson Jan 25 '18 at 16:43

You can split the sum and use

$$\sum_{n=1}^{\infty} r^n=\frac{r}{1-r}$$


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