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

Question

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?

Answer 1

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

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

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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.

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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.

Answer 2

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 $m\to\infty$.