Finding an upper-estimate for a sequence.

[Seydlitz]

Hi guys,

I'm on the verge of sandwiching this particular sequence but I need rather tight upper estimate to trap the limit to 1. I can only manage to get the sequence that converges to $e$ as the current upper estimate. Is it possible to get tighter bound than that?

<br /> \\<br /> 1 + \frac{1}{n^2} \le \left(1+\frac{1}{n^3}\right)^{n} \le \left(1+\frac{1}{n}\right)^n<br />

Alternatively, I can maybe show that the sequence is decreasing and that the limit infimum of the middle sequence is 1. But it might not be as simple as using sandwich theorem.

Thank You

P.S: In general, is there any inequality that shows the upper-estimate of $(1+x)^n$ in contrast to the Bernoulli's inequality that shows the lower-estimate of it?

 

[mfb]

You can compare it to $\left(1+\frac{1}{mn}\right)^n$ for every natural number m.

 

[pasmith]

Hi guys,

I'm on the verge of sandwiching this particular sequence but I need rather tight upper estimate to trap the limit to 1. I can only manage to get the sequence that converges to $e$ as the current upper estimate. Is it possible to get tighter bound than that?

<br /> \\<br /> 1 + \frac{1}{n^2} \le \left(1+\frac{1}{n^3}\right)^{n} \le \left(1+\frac{1}{n}\right)^n<br />

Alternatively, I can maybe show that the sequence is decreasing and that the limit infimum of the middle sequence is 1. But it might not be as simple as using sandwich theorem.

Alternatively, you can show by l'Hopital's Rule that \lim_{n \to \infty} n\log\left(1 + \frac1{n^3}\right) = \lim_{x \to 0^{+}} \frac{\log(1 + x^3)}{x} = 0.

 

[Erland]

If all you want is to show that

$\lim_{n\to \infty} (1+\frac 1{n^3})^n=1$,

then I wouldn't use sandwiching and estimates, but a direct approach:

$(1+\frac 1{n^3})^n=((1+\frac 1{n^3})^{n^3})^{1/n^2}\to e^0=1$, as $n\to\infty$.

 

[Seydlitz]

Ok guys thanks for the input. I'll try the sandwiching the guys first.