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Is $ \sum_{k=m+1}^{n} \frac{1}{2^k} < \frac{1}{2^m} $ true in general? Does it require induction on both $m$ and $n$?

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3 Answers 3

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$$1 = \sum_{k=1}^\infty \frac{1}{2^k} > \sum_{k=1}^{n-m} \frac{1}{2^k} = \sum_{k=m+1}^n \frac{1}{2^{k-m}}$$

Multiply both sides by $\frac{1}{2^m}$ and you are done.

More generally, if $0<z<1$ then:

$$\frac{z}{1-z} = \sum_{k=1}^\infty z^k > \sum_{k=1}^{n-m} z^k = \sum_{k=m+1}^n z^{k-m}$$

Multiply both sides by $z^m$ and you get:

$$\frac{z^{m+1}}{1-z} > \sum_{k=m+1}^n z^k$$

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This requires nothing more than the formula for the sum of geometric series. A geometric series starting with $A$ and with a common ratio $r$ has, after $\ell$ terms,

$$\sum_{i=0}^{\ell-1} A r^i = A \frac{1-r^{\ell}}{1-r}$$

Try putting in $A = 2^{-(m+1)}$ and $r=1/2$ and considering the limit as $\ell \to \infty$.

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Yes, this holds; you don't need induction; rather, note that the left-hand-side is made strictly greater by changing the $n$ to $\infty$, and after making this replacement the sum on the LHS is $\frac{1}{2^m}$.

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