How would you analytically prove that:(adsbygoogle = window.adsbygoogle || []).push({});

[itex] \displaystyle{\lim_{m\rightarrow ∞}} [/itex]([itex]\frac{1}{m}[/itex][itex]\sum[/itex][itex]^{n=1}_{m}[/itex][itex]\frac{1}{n}[/itex])= 0 ?

The way I did it, I just proved that the greatest lower bound of [itex]\frac{1}{m}[/itex] is 0, and so since the function is monotonically decreasing I proved that the limit of [itex]\frac{1}{m}[/itex] is 0, so anything multiplied by 0 (like the limit of [itex]\sum[/itex][itex]^{n=1}_{m}[/itex][itex]\frac{1}{n}[/itex]) must also be 0.

My professor agreed with me.

But now that I look at it again, obviously it can't be right, because the limit of [itex]\frac{1}{m}[/itex][itex]\sum[/itex][itex]^{n=1}_{m}[/itex][itex]\frac{1}{n}[/itex] as m goes to infinity is infinity, and 0 times infinity is indeterminate...

But my professor seems really smart and paid great attention to my work...is he wrong?

(My exact work was as follows):

[itex] \displaystyle{\lim_{m\rightarrow ∞}} [/itex]([itex]\frac{1}{m}[/itex][itex]\sum[/itex][itex]^{n=1}_{m}[/itex][itex]\frac{1}{n}[/itex]) = [itex] \displaystyle{\lim_{m\rightarrow ∞}} [/itex]([itex]\frac{1}{m}[/itex])*[itex] \displaystyle{\lim_{m\rightarrow ∞}} [/itex]([itex]\sum[/itex][itex]^{n=1}_{m}[/itex][itex]\frac{1}{n}[/itex])

Letbbe the greatest lower bound of [itex]\frac{1}{m}[/itex]. If b=0, then:

[itex]\frac{1}{m}[/itex]>0 for all m

Since b+ε cannot be a greatest lower bound, then:

0+ε>[itex]\frac{1}{N}[/itex] for some N

if m≥N, then f(N)≥[itex]\frac{1}{m}[/itex], since the function is monotonically decreasing. So:

0+ε>f(N)>[itex]\frac{1}{m}[/itex]>0

Therefore:

ε>[itex]\frac{1}{m}[/itex]>0, where ε>0 can be made arbitrarily close to 0.

Therefore, by the squeeze theorem, the limit of [itex]\frac{1}{m}[/itex] as m→∞ is 0.

Therefore,

[itex] \displaystyle{\lim_{m\rightarrow ∞}} [/itex]([itex]\frac{1}{m}[/itex])*[itex] \displaystyle{\lim_{m\rightarrow ∞}} [/itex]([itex]\sum[/itex][itex]^{n=1}_{m}[/itex][itex]\frac{1}{n}[/itex])=0

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# How to analytically prove a sum goes to zero

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