# Prove by induction that $\sum_{n=1}^\infty \frac{1}{2^n} = 1$

The title explains the problem fairly well; is there a way to prove by induction that $\sum_{n=1}^\infty \frac{1}{2^n} = 1$. If not are there other ways?

I have thought of showing it by rewriting the series so that. $$\sum_{n=1}^\infty \frac{1}{2^n} = 1 \implies \sum_{n=1}^\infty \frac{1}{2}(\frac{1}{2})^{n-1} = 1$$

And then from there conclude that it is a geometric series with the values $r = 1/2$ and $a=1/2$ thus $$\sum_{n=1}^\infty \frac{1}{2^n} = \frac{1/2}{1-1/2} = 1$$

This seems like kind of a vodoo proof, so i was wondering if its possible to do this by induction?

• It's a geometric series... Why do you even need induction?
– MSDG
Commented Sep 7, 2018 at 13:26
• Maybe induction for the partial sums? Commented Sep 7, 2018 at 13:31
• @Sobi Just for fun - to know if it's doable, or would be better than using the definiton of convergens for geometric series.
– user488081
Commented Sep 7, 2018 at 13:33
• I don't understand what variable would you take for induction? $\sum_{n=1}^\infty a_n$ is one object, not a sequence of objects. That is, $n$ already varies from $1$ to infinity, there are no steps here. Maybe you would like to prove formula for $n$-th partial sum by induction? That would work. Commented Sep 7, 2018 at 13:38

As was pointed out in the comments, induction is a method to prove statements of the form "For all natural numbers m, ... [some property about $m$] ...", and the statement you want to prove does not have that form. If you really want to use induction, I'd suggest using it to prove (by induction on $m$) that the partial sums of your series are $$\sum_{n=1}^m\frac1{2^n}=1-\frac1{2^m}.$$ Then finish the job by recalling that an infinite sum is defined as the limit of the finite partial sums and noting that $1-\frac1{2^m}\to1$ as $m\to\infty$.

Since the post got edited a few times, I'm not completely sure that this is what you were looking for. However, it is a geometric series:

Let $q$ s.t. $|q|<1$, then $$\sum_{n=0}^\infty q^n$$ converges and $\sum_{n=0}^\infty q^n=\frac{1}{1-q}$

Now, considering that

$$\frac{1}{2^n}=(\frac{1}{2})^n$$

we derive

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

I'm not really sure about how/where/on what to use induction here. I mean, you might use induction to establish prerequisites for a convergence criterion (not necessarily giving you the value as well), e.g. induction over partial sums to show some bound, etc.

However, in general, calculation of the value of a series can be (or often is) terrible. That's why convergence criteria like the classification of the geometric series, where you even get a formula for the value of the series under easy prerequisites are (to my knowledge) so rare that you just have to use them.

Note, that proving the criterion of geometric series can involve induction, so you might as well apply such a proof of it directly to the case of $q=1/2$:

To show the above formula for these geometric series, you usually proceed by finding a formula for the n-th partial sum

$$\sum_{k=0}^n q^k$$

which, for $q=1$, just amounts to $n+1$ for all $n\in\mathbb{N}$. For $q\neq 1$, you then can proof that

$$\sum_{k=0}^n q^k=\frac{1-q^{n+1}}{1-q}$$

for all $n\in\mathbb{N}$. I'm pretty sure that this can be shown by induction. So you could prove this, or the special case of $1/2$ just by induction (if possible, I didn't check this) and then use it to determine the formula (or value if $q=1/2$) for the series in a larger proof by evaluating the limit of the fraction for $n\to\infty$.

However, also here there is a much nicer proof (at least I think so): You have

$$(1-q)\sum_{k=0}^nq^k=\sum_{k=0}^nq^k-\sum_{k=0}^nq^{k+1}=q^0+\sum_{k=1}^nq^k-\sum_{k=0}^{n-1}q^{k+1}-q^{n+1}=\\1-q^{n+1}+\sum_{k=1}^nq^k-\sum_{k=1}^nq^k=1-q^{n+1}$$

Dividing by $(1-q)$, which is possible as $q\neq 1$, gives the expression.

• Sorry about the editing. You comment below the derivation helps, and was really what i was looking for. One question though. Would the derivation from your post be considered as a valid proof?
– user488081
Commented Sep 7, 2018 at 13:47
• @sn3jd3r Sure, but it is in very compact notation and also requires some theorems about series and limits, at least that you can pull finite sums out of a series. By the way, I've edited my answer to include some more information.
– blub
Commented Sep 7, 2018 at 13:55