# Limits: rate of convergence of Euler

1. Feb 12, 2009

### Knissp

1. The problem statement, all variables and given/known data
Prove that the rate of convergence for Euler's method (in solving the problem y'=y) is comparable to 1/n by showing that $$lim_{n\rightarrow infinity} \frac{e-(1+\frac{1}{n})^n}{1/n}$$.

2. Relevant equations
Hint: Use L'Hopital's Rule
If lim x-> infinity f(x)/g(x) is in indeterminate form, then it can be evaluated as lim x-> infinity f'(x)/g'(x).
Hint: Use Maclaurin expansion for ln(1+t).
ln(1+t) = t - t^2/2 + t^3/3 - t^4/4 + t^5/5 .......

3. The attempt at a solution

L'Hopital's Rule:
Differentiating gives
$$lim_{n\rightarrow infinity} \frac{\frac{1}{n+1} - ln(1 + \frac{1}{n})(\frac{n+1}{n})^n}{-1/n^2}$$

$$ln (1+1/n) = 1/n - 1/(2n^2) + 1/(3n^3) - 1/(4n^4) + 1/(5n^5) ...$$

Not sure where to go from here. When I plug in the series, what can I do? I tried multiplying the fraction by $$\frac{-n^2}{-n^2}$$ but I don't think that gets me anywhere. I could apply L'Hopital again, but that seems like a monstrous function to differentiate. Any ideas?

2. Feb 12, 2009

### Dick

I would say don't use l'Hopital if you are going to use series expansions anyway. (1+1/n)^n=exp(n*ln(1+1/n)). Use the series expansion on n*ln(1+1/n). Only keep 2 terms. Factor out the e. Now series expand exp. Only keep the terms that are going lead to terms with a power of n less than two.

3. Feb 13, 2009

### Knissp

Great! that works out to e/2. Thanks for the help! I wonder why they hinted at L'hopital though.