Proof: Summing an Infinite Series

[wyhwang7]

Hello! I have a midterm soon, and I'm doing some homework problems. I've stumbled upon a problem that I can't answer for the life of me. Could anybody give me a walkthrough/explanation for the following problem? Any help would be greatly appreciated.

Let Sn be the nth partial sum of the harmonic series
infinite
S = ?
n=1

(a) Verify the following inequality for n=1,2,3. Then prove it for general n.

1/(2^(n-1) + 1) + 1/(2^(n-1) + 2) + 1/(2^(n-1) + 3) + . . . + 1/2^n is greater than or equal to 1/2

(b) Prove that S diverges by showing that Sn is greater than or equal to 1 + n/2 for N=2^n

Hint: Break up Sn into n+1 sums of length 1,2,4,8,..., as in the following:

S of 2^3 = 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8)

Thank you!

 

[soroban]

Hello, wyhwang7!

This is a proof that the Harmonic Series is divergent.

\(\displaystyle S\text{ is the Harmonic Series: }\;S \;=\;\sum^{\infty}_{k=1}\frac{1}{k} \;=\;1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \hdots\)

\(\displaystyle \text{Let }S_n\text{ be the }n^{th}\text{ partial sum: } \;\sum^n_{k=1}\frac{1}{k} \;=\;1 + \frac{1}{2} + \frac{1}{3} + \hdots + \frac{1}{n}\)

\(\displaystyle \text{(a) Verify the following inequality for }n=1,2,3.\:\text{ Then prove it for general }n.\)

. . \(\displaystyle \frac{1}{2^{n-1} + 1} + \frac{1}{2^{n-1} + 2} + \frac{1}{2^{n-1} + 3} + \hdots + \frac{1}{2^n} \;\geq\;\frac{1}{2}\)

\(\displaystyle P(1) \:=\:\frac{1}{2^0+1} \:=\:\frac{1}{2} \:\leq \:\frac{1}{2}\;\hdots\;\text{ true!}\)

\(\displaystyle P(2) \:=\:\frac{1}{2^1+1} + \frac{1}{2^2+1} \:=\:\frac{1}{3} + \frac{1}{4} \:=\:\frac{7}{12} \;\geq \;\frac{1}{2}\;\hdots\;\text{true!}\)

\(\displaystyle P(3) \:=\:\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \:=\:\frac{1066}{1680}\;\geq\;\frac{1}{2} \;\hdots\;\text{true!}\)


\(\displaystyle \begin{array}{c}\text{Look at }P(2).\qquad\qquad\qquad \\ \text{We have 2 fractions, each } \geq \tfrac{1}{4} \\ \\[-3mm] \qquad \qquad\text{Their sum is: }\(2) \:\geq \:2\left(\tfrac{1}{4}\right) \:=\:\tfrac{1}{2} \end{array}\)

\(\displaystyle \begin{array}{c}\text{Look at }P(3).\qquad\qquad\qquad \\ \text{We have 4 fractions, each } \geq \tfrac{1}{8}\\ \\ [-3mm] \qquad\qquad\text{Their sum is: }\(3) \;\geq\:4\left(\tfrac{1}{8}\right) \:=\:\tfrac{1}{2} \end{array}\)

\(\displaystyle \begin{array}{c}\qquad\text{Look at }P(n) \;=\;\frac{1}{2^{n-1}+1} + \frac{1}{2^{n-1}+2} + \frac{1}{2^{n-1}+3} + \hdots + \frac{1}{2^n} \\ \\[-3mm] \text{We have: }2^{n-1}\text{ fractions, each } \geq \tfrac{1}{2^n}\qquadd\qquad\qquad\qquad\qquad \\ \\[-3mm] \text{Their sum is: }\(n) \;\geq\;\left(2^{n-1}\right)\left(\tfrac{1}{2^n}\right) \:=\:\tfrac{1}{2}\qquad\qquad \end{array}\)

\(\displaystyle \text{(b) Prove that }S\text{ diverges by showing that: }\:S_n \;\geq \;1 + \frac{n}{2}\:\text{ for }N=2^n\)

\(\displaystyle \text{Hint: Break up }S_n\text{ into }n+1 \text{ sums of length }1,2,4,8,\hdots\:\text{ as in the following:}\)

. . \(\displaystyle S(8) \;=\;1 + \tfrac{1}{2}+ \left(\tfrac{1}{3}+\tfrac{1}{4}\right)) + \left(\tfrac{1}{5}+\tfrac{1}{6}+\tfrac{1}{7}+\tfrac{1}{8}\right)\)

\(\displaystyle \text{We have: }\;S \;=\;1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \left(\frac{1}{9} + \frac{1}{10} + \hdots + \frac{1}{16}\right) + \hdots\)

\(\displaystyle \text{In part (a), we have shown that each group is parentheses is greater than or equal to }\tfrac{1}{2}\)


\(\displaystyle \text{Hence: }\:S \;\geq \;1 + \tfrac{1}{2} + \tfrac{1}{2} + \tfrac{1}{2} + \hdots\)

. . \(\displaystyle \text{Therefore, }S\text{ diverges.}\)