How can I prove an infinite sequence with induction

Given a infinite sequence that converges at 1:

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

How can I formally prove this using induction?

Normally I would go about showing a base case, for some value of $n$, to prove this is actually right, but this seems to be misleading.

Not sure what I am missing, but any pointer as to how to engage proving inifinite sequences with induction would be much appreciate, as I have found no helpful information so far.

My point however formally is to prove with induction that the sequence when the $\lim_{n \to \infty} \frac{1}{2^n} = 1$.

Answer

You can't do induction on a limit of an infinite sequence but you can on every finite sequence.

So You can prove that $\sum\limits_{n=1}^M \frac 1{2^n} = 1-\frac 1{2^M}$ by induction.[1]

And from that you can conclude $\sum\limits_{n=1}^\infty \frac 1{2^n}=\lim\limits_{M\to \infty}\sum\limits_{n=1}^M \frac 1{2^n} =\lim\limits_{M\to \infty}(1-\frac 1{2^M}) =1 -\lim\limits_{M\to \infty}\frac 1{2^M}$.

And we can prove $\lim\limits_{M\to \infty}\frac 1{2^M}=0$[2].

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[1]: Base case: $\sum\limits_{n=1}^1 \frac 1{2}^n = \frac 12 = 1 - \frac 12$.

Inductive step:

Assume $\sum\limits_{n=1}^k \frac 1{2^n} = 1-\frac 1{2^k}$ then

$\sum\limits_{n=1}^{k+1} \frac 1{2^n} = 1-\frac 1{2^k}+ \frac 1{2^{k+1}}=$

$1-(\frac 1{2^k}- \frac 1{2^{k+1}})=$

$1-(\frac 2{2^{k+1}}- \frac 1{2^{k+1}})=$

$1-(\frac {2-1}{2^{k+1}})=1-\frac 1{2^{k+1}}$

[2].... seems kind of weird to jump from natural number induction to analysis of limits but...

For any $\epsilon; 1> \epsilon > 0$ then $M = \frac 1\epsilon > 1$ and $n > \log_2 M$ then $2^n > M =\frac 1\epsilon$ and $0< \frac 1{2^n} < \epsilon$.