A tricky limit

[Shadow]

I came across the limit [math]\lim_{n -> \infty} n(\frac{1}{\sqrt{e}} - {(1 - \frac{1}{2n})}^{n})[/math], which converges to [math]\frac{1}{8\sqrt{e}}[/math] and I have no idea how to evaluate it. The only thing I could think of is to replace the inverse square root of e with something that converges to the same value, except that if I do that it changes the limit (unless I use [math]{(1 - \frac{1}{2n})}^{n - \frac{1}{4}}[/math], but I've only determined this experimentally with the help of WolframAlpha). Any ideas?

 

Also, not that I think it would make any difference in this particular case, but you're not supposed to use either L'Hospital or Taylor.

[md65536]

I came across the limit [math]\lim_{n -> \infty} n(\frac{1}{\sqrt{e}} - {(1 - \frac{1}{2n})}^{n})[/math], which converges to [math]\frac{1}{8\sqrt{e}}[/math] and I have no idea how to evaluate it. The only thing I could think of is to replace the inverse square root of e with something that converges to the same value, except that if I do that it changes the limit (unless I use [math]{(1 - \frac{1}{2n})}^{n - \frac{1}{4}}[/math], but I've only determined this experimentally with the help of WolframAlpha). Any ideas?

 

Also, not that I think it would make any difference in this particular case, but you're not supposed to use either L'Hospital or Taylor.

 

I'm not a mathematician and I don't know how to do this, but...

 

Perhaps the fact that e^z can be defined as the limit of (1 + z/N)^N as N approaches infinity[1] provides a way of solving it.

With z=-1/2, you get [math] \frac{1}{\sqrt{e}} = \lim_{n -> \infty} (1 - \frac{1}{2n})^n[/math].

 

Then if you pick apart why e can be defined that way, it might give you the answer.

 

 

References:

[1] under animated gif at http://en.wikipedia....er%27s_identity

Edited January 21, 2012 by md65536

[mathematic]

expand 1/√e into a power series and (1 - 1/(2n))^n using the binomial. Subtract and the first two terms in both expansions cancel each other, while the others will have something with 1/n. Multiply by n and see what you get.

[Shadow]

I'm not a mathematician and I don't know how to do this, but...

 

Perhaps the fact that e^z can be defined as the limit of (1 + z/N)^N as N approaches infinity[1] provides a way of solving it.

With z=-1/2, you get [math] \frac{1}{\sqrt{e}} = \lim_{n -> \infty} (1 - \frac{1}{2n})^n[/math].

 

Then if you pick apart why e can be defined that way, it might give you the answer.

 

 

References:

[1] under animated gif at http://en.wikipedia....er%27s_identity

 

That's exactly what I tried to do, but as I said, if I replace the inverse square root with something that converges to it, it almost always changes the result of the limit. For example, if I used what you suggest the result would be trivially zero and if I used say [math]\frac{1}{\sqrt{e}} = \lim_{n -> \infty} (1 + \frac{1}{n})^{-\frac{n}{2}}[/math], the limit would converge to [math]\frac{3}{8\sqrt{e}}[/math] instead of [math]\frac{1}{8\sqrt{e}}[/math].

 

 

expand 1/√e into a power series and (1 - 1/(2n))^n using the binomial. Subtract and the first two terms in both expansions cancel each other, while the others will have something with 1/n. Multiply by n and see what you get.

 

Thanks. I'm not that well versed in power series yet, so I'm not too sure how to go from here:

 

[math]\frac{1}{\sqrt{e}} = e^{-\frac{1}{2}} =\sum_{k=0}^\infty \frac{(-1)^k}{2^k(k!)}[/math]

 

I can see that the first terms cancel out, but how do I get anything containing n (or 1/n) out of the power series? The only thing that occurs to me is to rewrite the power series as [math]\lim_{n->\infty} \sum_{k=0}^n ...[/math], but the closed form for [math]\sum_{k=0}^n ...[/math] contains gamma functions, which are beyond my level of comprehension and more importantly the inverse square root of e, which just gets us back were we started.

 

Also, neither L'Hospital's rule nor power series should be used. Frankly, given the fact that this is just a random exercise a friend sent me I'll settle for being able to solve it with the two, but if anyone sees an alternate, non-L'Hospital/power series method, it would be much apreciated.

Edited January 22, 2012 by Shadow

[mathematic]

I guess I didn't make my point clear enough. (1-1/(2n))^n expanded by the binomial theorem contains terms in n. These terms would be matched term by term with the terms in the power series for 1/√e and divided by n. The left over terms of the power series get very small even when multiplied by n. So examine the n+1 terms of the difference and see how it develops.

[md65536]

That's exactly what I tried to do, but as I said, if I replace the inverse square root with something that converges to it, it almost always changes the result of the limit.

Well that makes sense, because inside the parenthesis of your limit you have something that approaches 0, and you're multiplying it by something that approaches infinity. You can't just change the thing that approaches 0, even slightly, and expect the limit to evaluate to the same thing.

 

Basically you have two things that converge to the same value, and you're subtracting them.

 

The limits of [math]\frac{1}{\sqrt{e}}[/math] and [math](1 - \frac{1}{2n})^{n}[/math] are the same, but for a finite n they differ by a specific amount which you're trying to find (or were given, really), which is proportional to 1/n. If you pull out one of those values and put in a different value, even if it converges on the same value, it will differ by some other specific amount for a given n.

 

 

Anyway that doesn't tell you how to do it. mathematic tells you how. What does it look like when you expand (1-1/(2n))^n by the binomial theorem? (I don't know.)

Edited January 23, 2012 by md65536

[Shadow]

Gotcha. Okay, so when I expand (1-1/2n)^n I get

-(1/2)

+(-1 + n)/(8 n)

-(((-2 + n) (-1 + n))/(48 n^2))

+((-3 + n) (-2 + n) (-1 + n))/(384 n^3)

-(((-4 + n) (-3 + n) (-2 + n) (-1 + n))/(3840 n^4)

.

.

.

 

and when I expand the series I get

-(1/2)

+1/8

-(1/48)

+1/384

-(1/3840)

.

.

.

 

The denominators agree up to a power of n, so when we find the common denominator and do the difference, we get rid of the highest power of n in the numerator (which is the same power as the highest power in the denominator), but when we multiply by the n in the limit we get a fraction that has the same highest power of n in both the numerator and denominator, with the coefficient of the highest power in the denominator is always one, and the coefficient of the highest power in the numerator of the kth fraction should be [math]\frac{(-1)^{k+1}*k*(k+1)}{2^{k+1}*k!}[/math], which is what the fraction will converge to. Unfortunately, a sum of these terms converges to 3/8*Sqrt[e], instead of 1/8*Sqrt[e], so I must've made a mistake somewhere, but I can't figure it out.

Also, how would one go about showing that the sum of [math]\frac{(-1)^{k+1}*k*(k+1)}{2^{k+1}*k!}[/math] converges to 3/8*Sqrt[e]?

[mathematic]

I'll let you do the details. However k(k+1)/k! = 1/(k-2)! + 2/(k-1)!. These will give expressions which will look something like the series for 1/√e.

[Shadow]

Okay, I think I could probably work that out on my own. But I'm still confused as to why I got 3/8*Sqrt[e] instead of 1/8*Sqrt[e].

[mathematic]

Okay, I think I could probably work that out on my own. But I'm still confused as to why I got 3/8*Sqrt[e] instead of 1/8*Sqrt[e].

Could you have a sign error? I can't tell unless I see the actual work.

[Shadow]

After a bit of playing around with WA I discovered that the k*(k+1) should actually be k^2 (or at least the sum works out this way). So who knows, maybe it was a sign error somewhere, maybe something else. But it's irrelevant, I was only after the method of solving this limit. Many thanks for your help, it was much appreciated