Proof by Induction: Tricky Sum

Question

Let $T_k = 1 + (1/2) + (1/3) + (1/4) ... + (1/k)$ for any $k \in\ \mathbb{N}$

Show that for $k = 2^n$ for any $n \in\ \mathbb{N}$, we have $T_k \geq 1 + (n/2)$

I have been working on this problem for a few hours now. I've shown that it works for P(1):

$$P(1): 1 + (1/2) \geq 1 + (1/2) ✓$$

Now I need to show $P(n+1)$. This is my setup:

$$\sum_{m=0}^{2^{n+1}} \frac{1}{2^n} = (\sum_{m=0}^{2^n} \frac{1}{2^n}) +\frac{1}{2^{n+1}} $$

which is:

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

I'm not sure where to go from here. I know I need to show that this is greater than or equal to $(1+\frac{n+1}{2})$

I note $1 \geq 1/2^0$; $1/2 \geq 1/2^1$; and $1/3,1/4 \geq 1/2^2$. Which subsequent fractions are $ \geq 1/2^3$? — Eric Towers, Sep 15 '20 at 22:11
By the way, you may want to consider a comparison of $T_k$ to the integral of $\frac1x$ — Maximilian Janisch, Sep 15 '20 at 22:15
The summand should be $1/m$, not $1/2^n$, and the sum should start at $m=1$. — RobPratt, Sep 15 '20 at 22:17

Tortar · Answer 1 · 2020-09-15T22:37:53.393

0

$$\sum_{m=1}^{2^{n+1}} \frac{1}{m} = \sum_{m=1}^{2^n} \frac{1}{m} +\sum_{m=2^n+1}^{2^{n+1}} \frac{1}{m} \geq 1+\frac{n}{2}+\sum_{m=2^n+1}^{2^{n+1}} \frac{1}{m}\geq 1+\frac{n+1}{2} $$

Indeed $\sum_{m=2^n+1}^{2^{n+1}} \frac{1}{m}> \sum_{m=2^n+1}^{2^{n+1}} \frac{1}{2^{n+1}} = 1/2 $

edited Sep 15 '20 at 22:37

answered Sep 15 '20 at 22:23

Tortar

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Proof by Induction: Tricky Sum

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