# Proof using Taylor's theorem

Use Taylor's theorem to prove that $\displaystyle\lim_{n \to \infty} n \ln\left(1+\frac{1}{n}\right)=1$

I don't understand how to apply Taylor's theorem to a limit, especially one with a product of two functions...

• try using taylor series expansion for $\ln (1+x)$ – r9m Mar 3 '14 at 4:48
• Ok, I understand this, but how do I multiply it by n and take the limit? – kiwifruit Mar 3 '14 at 4:52
• see that the limit exists .. $\frac{1}{n+1}< \ln (1+\frac{1}{n})< \frac{1}{n}$, we are only applying the Taylor series to evaluate the limit at the point :) – r9m Mar 3 '14 at 4:55
• Duplicate of math.stackexchange.com/q/698331/18398 – Joel Reyes Noche Mar 4 '14 at 1:11
• @JoelReyesNoche It is a duplicate, but this is the original version of the question. The other version has four votes for closure already, so it should be marked as duplicate soon. – user61527 Mar 4 '14 at 4:08

As $\displaystyle n\to\infty,\frac1n\to0\implies \left|\frac1n\right|<1$

Using $\#19$ of this , $$\ln\left(1+\frac1n\right)=\frac1n-\frac1{n^2}+\frac1{n^3}-\cdots$$

• Ok, I understand this, but how do I multiply it by n and take the limit? – kiwifruit Mar 3 '14 at 4:51
• @kiwifruit, $$\lim_{n\to\infty}\frac nn=?$$ and $$\lim_{n\to\infty}\frac n{n^k}=?$$ for integer $k>1$ – lab bhattacharjee Mar 3 '14 at 4:52
• Is it correct that $\lim_{n \to \infty} \frac{n}{n^k} = \frac {lim_{n \to \infty} n}{lim_{n \to \infty} n^k}?$ – kiwifruit Mar 3 '14 at 4:54
• @kiwifruit No. Since $\lim_{n\to\infty} n$ and $\lim_{n\to\infty} n^k$ are not converge, so you do not apply the theorem $\lim_n a_nb_n = \lim a_n \cdot \lim b_n$. – Hanul Jeon Mar 3 '14 at 5:16

By Taylor's theorem, we have $$\ln(1+t) = t + o(t)$$ as $t \to 0$. Substituting $t = \frac{1}{n}$, we get, as $n \to \infty$, $$\ln(1 + \frac{1}{n}) = \frac{1}{n} + o(\frac{1}{n})$$

Thus, we have $$n \ln(1 + \frac{1}{n}) = n \left[\frac{1}{n} + o(\frac{1}{n})\right] = 1 + n \cdot o(\frac{1}{n}) = 1 + \frac{o(\frac{1}{n})}{\frac{1}{n}} \to 1 + 0 = 1$$ as $n \to \infty$

Recall that $$\ln(1+x)=\sum_{k=1}^\infty(-1)^{k+1}\frac{x^k}{k}$$ for $|x|<1$. Then \begin{align*} \lim_{n\to\infty}n\ln\left(1+\frac{1}{n}\right) &=\lim_{n\to\infty}n\sum_{k=1}^\infty(-1)^{k+1}\frac{1}{n^k k} \\ &=\lim_{n\to\infty}\sum_{k=1}^\infty(-1)^{k+1}\frac{1}{n^{k-1} k} \\ &=\lim_{n\to\infty}\left\{1+\sum_{k=2}^\infty(-1)^{k+1}\frac{1}{n^{k-1} k}\right\} \\ &=1+\sum_{k=2}^\infty(-1)^{k+1}\lim_{n\to\infty}\frac{1}{n^{k-1} k} \\ &= 1 \end{align*}