Limit of sequence

[twoflower]

Hi all,

I can't move with this limit:


\lim_{n \rightarrow \infty} n \left( \left( 1 + \frac{1}{n} \right)^{n} - e \right)


Could someone help me please? Or some hint...

But no l'Hospital please.

Thank you

[e(ho0n3]

Find out what


\lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right)^{n}


is. It should all be clear from then on.

[twoflower]

Well I know it is equal to e, but then


\lim_{n \rightarrow \infty} n \left( e - e \right) = \lim_{n \rightarrow \infty} n \left ( 0 \right) = 0.\infty


Which is undefined...

[shmoe]

Are you opposed to using the series expansions of log(1+x) and e^x? You can re-write the (1+1/n)^n term as e^(n*log(1+1/n)). Use the Taylor series for log(1+x), at least 2 terms +remainder. Then you can factor out an e and use the series for e^x, (2 terms+remainder here as well), and that will do it.

If you are opposed to using series but have learned them, it's probably possible to modify the above into a squeeze limit type of proof. If you haven't learned Taylor series yet, I'll have to think of a more elementary method, though it will probably be a thinly disguised version of the above.

[twoflower]

Thank you shmoe, maybe this limit is really supposed to be solved using Taylor series, because it is from sample calculus test we're gonna take, but as I see in sylabus, we'll learn Taylor series before the test. It confused me, because the sequences are already behind us and I thought I should already be able to solve any limit of sequence....

[stunner5000pt]

Well I know it is equal to e, but then


\lim_{n \rightarrow \infty} n \left( e - e \right) = \lim_{n \rightarrow \infty} n \left ( 0 \right) = 0.\infty


Which is undefined...

you cannot sub in the value for e in between the limit like that

[stunner5000pt]

i wonder if this will work though

split the limit so you have the limits of n(1-1/n)^n and n e subtracted from each other.

now the limit of n (1-1/n)^n is infinity while the limit of n E is also infinity

however the first one converges to e more slowly that E already does

so n E >n(1-1/n)^n and the limit is negative infinity

[twoflower]

The limit should be -e/2...

[stunner5000pt]

The limit should be -e/2...

check what you entered into maple/mathematica/matlab... it isn't -e/2, it is -infinity

[shmoe]

i wonder if this will work though

split the limit so you have the limits of n(1-1/n)^n and n e subtracted from each other.

now the limit of n (1-1/n)^n is infinity while the limit of n E is also infinity

however the first one converges to e more slowly that E already does

so n E >n(1-1/n)^n and the limit is negative infinity

No, this won't work. You're essentially saying infinity-infinity=-infinity because the first infinity is getting there slower. This is bunk. An indeterminate infinity-infinity limit form can potentially equal anything we like(-e/2 is correct in this case, done by hand with the method I suggested).

twoflower-series is likely what's intended then. I can't see another way that isn't unnecessarily complicated.

[NateTG]

Hi all,

I can't move with this limit:


\lim_{n \rightarrow \infty} n \left( \left( 1 + \frac{1}{n} \right)^{n} - e \right)


Could someone help me please? Or some hint...


If you can show that
-\frac{e}{2n} - \epsilon_1(n) \leq \left( 1 + \frac{1}{n} \right)^{n} - e \leq -\frac{e}{2n} + \epsilon_2(n)

Where \epsilon(n) is some expression that goes to zero faster than
\frac{1}{n} then you're set, so perhaps you should look at the convergence of \left( 1 + \frac{1}{n} \right)^{n}

[twoflower]

check what you entered into maple/mathematica/matlab... it isn't -e/2, it is -infinity

I tried it now. Maple gives -e/2

[shmoe]

Nate, do you have a simple method in mind for those inequalities?

The obvious approach to me involves some bounds on log(1+x) and e^x that I would prove using series.

[twoflower]

Nate, do you have a simple method in mind for those inequalities?

The obvious approach to me involves some bounds on log(1+x) and e^x that I would prove using series.

Ok, so you say guys that the most natural way to find this limit is to use Taylor series. If it's right, I will skip this one. Or does anyone have simplier approach to show?