# How to find the limit of series? (What should I know?)

There is a couple of limits that I failed to find:

$$\lim_{n\to\infty}\frac 1 2 + \frac 1 4 + \frac 1 {8} + \cdots + \frac {1}{2^{n}}$$

and

$$\lim_{n\to\infty}1 - \frac 1 3 + \frac 1 9 - \frac 1 {27} + \cdots + \frac {(-1)^{n-1}}{3^{n-1}}$$

There is no problem to calculate a limit using Wolfram Alfa or something like that. But what I am interested in is the method, not just a concrete result.

So my questions are:

• What should I do when I need a limit of infinite sum? (are there any rules of thumb?)
• What theorems or topics from calculus should I know to solve these problems better?

I am new to math and will appreciate any help. Thank you!

• Hint: Geometric Series! – Qi Zhu Oct 29 '17 at 19:24
• I am also new to Mathematics on Stack Exchange, so if you see that I can improve my question, I would appreciate if you left a comment – Nikita Hismatov Oct 29 '17 at 19:24
• Demidovich!!! I've used it in 1977 to prepare Calculus I and after I continued to use it to give private classes to a lot of people, including the girl who became my wife... – Raffaele Oct 29 '17 at 20:58
• Writing the first in binary will result in something interesting. – chx Oct 29 '17 at 21:00
• @NikitaHismatov Once I asked a professor of mine about this. He replied: "Do you think we have a panacea to calculate limits?!" - As it often happens in mathematics, we have methods for some cases but you can always find harder limits for which the previously developed methods do not work. I once thought about the existence of the following limit and asked here, people were able to compute the limit but I'm not sure I really understand what they did. – Billy Rubina Oct 30 '17 at 6:34

$$\color{red}{1+}\frac 1 2 + \frac 1 4 + \frac 1 {8} + \ldots + \frac {1}{2^{n}}$$ Is an example of $$1+x+x^2+x^3+\ldots+x^n$$ Where the ratio between a number $a_n$ and the previous $a_{n-1}$ is constant and is $x$. This is the sum of a geometric progression and it's quite easy to see that its value is $$\frac{1-x^{n+1}}{1-x}$$ When $n\to\infty$ the sum converges only if $|x|<1$ because if so $x^{n+1}\to 0$ and the sum is $$\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$$ In your first example $x=\frac12$ and index $n$ starts from $n=1$ so the sum is $$\sum_{n=1}^{\infty}\left(\frac12\right)^n=\frac{1}{1-\frac12}-1=\color{red}{1}$$ The second one is $$\lim_{n\to\infty}1 - \frac 1 3 + \frac 1 9 - \frac 1 {27} + \cdots + \frac {(-1)^{n}}{3^{n}}=\sum_{n=0}^{\infty}\left(-\frac13\right)^n=\frac{1}{1-\left(-\frac13\right)}=\color{red}{\frac34}$$

Hope this is useful

Hint:

All you have to know is the sum of the $n$ first terms of a geometric series, which is a formula from high school: $$1+q+q^2+\dots +q^n=\frac{1-q^{n+1}}{1-q}\qquad (q\ne 1),$$from which we can deduce: $$q^r+q^{r+1}+\dots +q^n=q^r(1+q+\dots+q^{n-r})=q^r\frac{1-q^{n-r+1}}{1-q}=\frac{q^r-q^{n+1}}{1-q}.$$ If $\lvert q\rvert<1$, these sums have limits $\;\dfrac1{1-q}\;$ and $\;\dfrac{q^r}{1-q}\;$ respectively.

What should I do when I need a limit of infinite sum? (are there any rules of thumb?)

In general - no there aren't such rules. There are specific types of series for which it is known how to compute the limit (like the geometric series). There are way more recepies for figuring out whether a series converges or not (finding the limit is much harder). But also here the cookbook is limited. As others pointed out already, your two series are geometric series, who's limit can be computed easily.

Well, that first one should be an instance of a theorem you should definitely put in your list of useful theorems:

If $|a| < 1$:

$$\sum_{i=1}^{\infty} \ a^i = \frac{a}{1-a}$$

And the second one is an instance of the closely related:

If $|a| < 1$:

$$\sum_{i=0}^{\infty} \ a^i = \frac{1}{1-a}$$

If you don't want to remember both, it's ok to remember just one, because the other one can easily be derived from it, e.g:

$$\sum_{i=0}^{\infty} \ a^i = a^0 + \sum_{i=1}^{\infty} \ a^i = 1 + \frac{a}{1-a} = \frac{1-a}{1-a} + \frac{a}{1-a} = \frac{1-a+a}{1-a}= \frac{1}{1-a}$$

• Or, perhaps more simply, $$\sum_{i=1}^{\infty}a^i = a\sum_{i=0}^{\infty}a^i =a\left( \frac{1}{1-a}\right)$$ – Bungo Oct 29 '17 at 19:52
• @Bungo Yes, nice! I always forget about that trick :( – Bram28 Oct 29 '17 at 19:55

the other answers are correct but if you want to see a more "visualize" way(it is actually the same way but you can see it more clearly) $$S=\lim_{n\to\infty}\sum_{i=1}^n\frac {1}{2^{i}}=\frac 1 2 + \frac 1 4 + \frac 1 {8} + \cdots\\2S=\lim_{n\to\infty}\sum_{i=0}^n\frac {1}{2^{i}}=\frac 1 1 + \frac 1 2 + \frac 1 {4} + \cdots=1+\lim_{n\to\infty}\sum_{i=1}^n\frac {1}{2^{i}}=1+S\\2S-S=\left(1+S\right)-S=1$$ and $$A=\lim_{n\to\infty}\sum_{i=1}^n\frac {(-1)^{i-1}}{3^{i-1}}=1 - \frac 1 3 + \frac 1 9 - \frac 1 {27} + \cdots \\ 3A=\lim_{n\to\infty}\sum_{i=1}^n\frac {(-1)^{i-1}}{3^{i-2}}=3 - \frac 3 3 + \frac 3 9 - \frac 3 {27} + \cdots\\=3-\left(1 - \frac 1 3 + \frac 1 9 - \frac 1 {27} + \cdots\right)=3-A\\3-A=3A\implies3=4A\implies\frac34=A$$

Note that you have the sums $\sum_{k = 1}^{\infty} \frac{1}{2^k}$ and $\sum_{k=0}^{\infty} \left(-\frac{1}{3} \right)^k$. Now look up geometric series and try to finish these on your own :)

In case of infinite Geometric Progression,

$$sum = \frac{a}{1-r}$$

you can also derive this from the normal formula using, $n\to \infty$

The thing to know here is that $$|r| <1$$

To explain this, if |r|<1 , ar < a similarly $ar^2 < ar$ and each next term will keep getting smaller and smaller and as $n\to \infty$, $$a_n = ar^{n-1} \to 0$$ therefore, we can successfully find a finite number to which this sequence will converge to as the later on terms are so smaller than the initial ones that they won't even contribute to the sum.

Although if |r|>1 each term will be greater than the previous one, so we will never be able to sum this sequence.

The sums you mention are both examples of geometric series. There are no general ways to sum arbitrary sequences - they typically require some inspiration. Geometric series are often proven geometrically, but the algebraic proof is illustrative.

$$\sum_{n=0}^{m} a^n = S_{m}, a S_{m} = \sum_{n=1}^m a^n = S_{m+1} - 1$$

$$S_m - a S_m = 1 - a^{m+1}$$

$$S_{m} = \frac{1 - a^{m+1}}{1 - a}$$

The other series is relatively straightforward if you know what your goal is:

$$\sum_{n=0}^\infty \frac{(-1)^n}{3^n} = \sum_{n=0}^\infty \frac{1}{9^n} - \frac{1}{3}\sum_{n=0}^\infty \frac{1}{9^n} = \frac{2}{3} \sum_{n=0}^\infty \frac{1}{9^n}$$

They are sums of geometric progressions.

The first it's $$\frac{\frac{1}{2}}{1-\frac{1}{2}}=1.$$

The second it's $$\frac{1}{1+\frac{1}{3}}=\frac{3}{4}.$$

• I downvoted this answer because it does not address the OP's note that "what I am interested in is the method, not just a concrete result." Naming the series is helpful, but something more—like explaining a closed form for the partial sums, or showing that the limit exists and then showing what it must be—is in order. – wchargin Oct 29 '17 at 22:14