# 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

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#### 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

Science Advisor
Homework Helper
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

twoflower said:
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

twoflower said:
The limit should be -e/2...
check what you entered into maple/mathematica/matlab... it isn't -e/2, it is -infinity

#### shmoe

Science Advisor
Homework Helper
stunner5000pt said:
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

Science Advisor
Homework Helper
twoflower said:
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

stunner5000pt said:
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

Science Advisor
Homework Helper
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

shmoe said:
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?

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