Find a limit

[hank]

By computing H_2n and H_n for some large values of n (or otherwise) guess the value of

the limit as n -> infinity (H_2n - H_n.)

Where H_2n is the harmonic series 1/2n and
H_n is the harmonic series 1/n.

What I tried to do was to use an integral approximation for both of them, so I take the limit of
1 + .5lnn - (1 + lnn) = - .5lnn.

However, this seems to be infinity, which I do not think is correct.

Help?

 

[mmm4444bot]

Can you do the subtraction first, and take the limit of -1 over 2n, as n goes to infinity?

 

[galactus]

$\displaystyle \frac{1}{2n}-\frac{1}{n}=\frac{-1}{2n}$

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

This is just the Harmonic series, only negative.

This does not converge, as we know.

But, did you try relating it to the gamma function.

$\displaystyle {\gamma}=\lim_{n\to {\infty}}\sum_{k=1}^{n}\frac{1}{k}-ln(n+1)$

$\displaystyle {\gamma}\approx .577$

 

[hank]

Thanks for the replies.

How would I go about relating it to gamma?

I've seen language like that used in other problems, but I'm not clear what that means.
I think we may have skipped over material related to that in earlier courses, or it's been so long since I've seen it I forget.

Does that just mean the limit is about .577?

 

[BigGlenntheHeavy]

$\displaystyle Given: \ H_{2n} \ = \ \frac{1}{2n}, \ H_n \ = \ \frac{1}{n}, \ Find \ \lim_{n\to\infty}(H_{2n}-H_{n})$

$\displaystyle \lim_{n\to\infty}(H_{2n}-H_{n}) \ = \ \lim_{n\to\infty}\bigg(\frac{1}{2n}-\frac{1}{n}\bigg) \ = \ \lim_{n\to\infty}\bigg(\frac{n-2n}{2n^{2}}\bigg)$

\(\displaystyle = \ \lim_{n\to\infty}\frac{-n}{2n^{2}} \ = \ \lim_{n\to\infty}\frac{-1}{2n}} \ = \ - \frac{1}{2}\lim_{n\to\infty}\frac{1}{n} \ = \ -\frac{1}{2}(0) \ = \ 0\)

$\displaystyle Am \ I \ missing \ something?$

 

[hank]

Yes.
H_2n and H_n are harmonic series, not just regular functions.
That's what has me confused. =(
So H_2n = 1/2 + 1/4 + 1/6 + ...
H_n = 1 + 1/2 + 1/3 + 1/4 + ...

 

[galactus]

As i see it, the Harmonic series is a limit of a sum. It is not just $\displaystyle \lim_{n\to {\infty}}\frac{1}{n}$.

It is the sum of the first n reciprocals as n-->infinity. See what I am getting at?.

At least, I think that is what Hank means. Maybe I am wrong.

$\displaystyle =\frac{-1}{2}\sum_{k=1}^{n}\frac{1}{k}=\frac{-1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+.........+\frac{1}{n}\right)$

$\displaystyle -\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+....................\frac{1}{n}\right)$

But, as n gets larger and larger, we get the gamma function:

$\displaystyle \frac{-1}{2}\lim_{n\to {\infty}}\sum_{k=1}^{n}=\frac{-1}{2}\left({\gamma}+ln(n+1)\right)$

Try a value of n and see. As it gets larger and larger, it approaches said result.

$\displaystyle \frac{-1}{2}\sum_{k=1}^{100}=-2.5936887588......$

Now, plug into $\displaystyle \frac{-1}{2}\left(.577+ln(101)\right)=-2.5960602584...$

Just a fun result from the gamma function.

As n gets larger and larger, we head into negative territory and the limit of the sum is $\displaystyle -{\infty}$

Which stands to reason, since we know the Harmonic series diverges. This one has a negative tacked on the front , so it 'approaches' negative infinity.

I do not like to say anything 'approaches' infinity. That is why I wrapped it in quotes.

Also, have you heard of the Psi function?. Technically, we get $\displaystyle \frac{-1}{2}\lim_{n\to {\infty}}\sum_{k=1}^{n}\frac{1}{k}=\frac{-1}{2}\lim_{n\to {\infty}}({\Psi}(n+1)+{\gamma})={-\infty}$

 

[hank]

I haven't heard of the psi function, but in any case, it looks like my original answer was correct.
The limit is -infinity?

 

[BigGlenntheHeavy]

$\displaystyle \lim_{n\to\infty}-\frac{1}{n} \ = \ 0$

$\displaystyle However, \ -\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{k} \ = \ -\infty, \ since \ \frac{1}{k} \ diverges \ (very, \ very, \ very, \ slowly \ to \ infinity).$