Prove inequality $0< \frac{1}{n} \ -\ln\left( 1+\frac{1}{n} \right) < \frac{1}{2n^{2}}$ using Taylor polynomial

Question

Proof with Taylor polynomial for $\displaystyle\\f( x)=\ln( 1+x)$ \begin{equation*} \forall \ n\in \mathbb{N} \ ,\ n\geq 2,\qquad 0< \frac{1}{n} \ -\ln\left( 1+\frac{1}{n} \right) < \frac{1}{2n^{2}} \end{equation*}

I calculated the $\displaystyle f( x) =T_{n}( x) +R_{n}( x) \ =\sum ^{n}_{k=1}( -1)^{k+1}\frac{x^{k}}{k} \ +\frac{( -1)^{n+2}}{n+1}\frac{x^{n+1}}{( 1+c)^{n+1}}$ $ $ $\displaystyle \begin{array}{{>{\displaystyle}l}} \text{for some }c\ :\ \ 0< c< x\\ \ \ \end{array}$

but I don't know how that supposed to help me , if someone could give me clue what I'm supposed to do that would be awesome

Answer 1

You may recall that $\int_{0}^{x}\frac{du}{1+u}$ is an equivalent definition of $\log(1+x)$. It follows that $$ \frac{1}{n}-\log\left(1+\frac{1}{n}\right) = \int_{0}^{\frac{1}{n}}\frac{u}{u+1}\,du $$ is positive as the integral of a positive function and bounded by $$ \int_{0}^{\frac{1}{n}}u\,du = \frac{1}{2n^2}.$$

Answer 2

Express $-\ln(1+\frac1n)$ in its Taylor polynomial form,

$$-\ln(1+\frac1n) =-\frac1n+\frac1{2n^2}-\frac1{3n^3}+\frac1{4n^4}...$$

Then,

$$\frac 1n - \ln(1+\frac1n) = \frac1{2n^2}-\frac1{3n^3}+\frac1{4n^4}-...$$ $$=\int_0^{\frac1n}(x-x^2+x^3-...)dx=\int_0^{\frac1n}\frac x{1+x}dx>0\tag 1$$

Similarly,

$$\frac 1n - \ln(1+\frac1n) - \frac1{2n^2} = -\frac1{3n^3}+\frac1{4n^4}-\frac1{5n^5}...$$ $$=\int_0^{\frac1n}(-x^2+x^3-x^4...)dx=\int_0^{\frac1n}\frac {-x^2}{1+x}dx<0\tag 2$$

Combine (1) and (2) to arrive at the inequalities,

$$0< \frac{1}{n} \ -\ln( 1+\frac{1}{n}) < \frac{1}{2n^{2}} $$

Answer 3

All you need is the first-degree Taylor polynomial with remainder (i.e., your formula with $n=1$ — although using $n$ is not so good because so many $n$'s appear in your problem). You were on the right track, but you got lost in symbols. So you have $$\ln(1+x) = x - \frac12\cdot\frac1{(1+c)^2} x^2 \quad\text{for some $c$ between $0$ and $x$}.$$ In particular, for $x>0$ (which forces $c>0$) we have $$0<x-\ln(1+x) = \frac12\cdot\frac1{(1+c)^2}x^2 < \frac12 x^2.$$ Substitute $x=1/n$ and you're done.

Answer 4

Note that $${1\over n}-\ln \left(1+{1\over n}\right){=\sum_{k=2}^{\infty} {(-1)^k\over kn^k}\\=\sum_{k=1}^{\infty} {1\over 2kn^{2k}}-{1\over (2k+1)n^{2k+1}}}$$which is always positive since for $n>1$$${2kn^{2k}}<{(2k+1)n^{2k+1}}$$ This proves the left inequality.

Also$${1\over 2n^2}-{1\over n}+\ln \left(1+{1\over n}\right){=\sum_{k=3}^{\infty} {(-1)^{k+1}\over kn^k}\\=\sum_{k=1}^{\infty} {1\over (2k+1)n^{2k+1}}-{1\over (2k+2)n^{2k+2}}}$$which is again always positive for $n>1$ with the same reason as before. This proves the right inequality.