help with sum of infinite series, stuck in problem

Question

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...

Answer 1

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

Answer 2

You can split the sum and use

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

Answer 3

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!

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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}

Answer 4

Thanks for the question. I'd say you could try writing the series (let $S=\sum_{n \ge 1}\frac{2^{n+1}+1}{3^n}$) as $S=U_1+U_2$, where $U_1=\sum_{n \ge 1}\frac{2^{n+1}}{3^n}$ and $U_2=\sum_{n \ge 1}\frac{1}{3^n}$. Let $U_1$ be re-written as $2\sum_{n \ge 1}\frac{2^{n}}{3^n}$. Now, as already mentioned by users before me, when $|x|<1$, can you derive the expression for $\sum_{n \ge 1}x^n$? This would be typically high school level algebra. This would give both $U_1$ and $U_2$. Then you can add the results.