Prove $\forall k\geq 4,\log(1+x_k)-x_k\leq {-1\over6k}$ where $x_k={(-1)^k\over\sqrt k}$

Question

The question:

Prove $\forall k\geq 4,\log(1+x_k)-x_k\leq {-1\over6k}$ where $x_k={(-1)^k\over\sqrt k}$.

The above inequality holds iff $$\begin{align} &\log(1+x_k)\leq x_k-{1\over 6k}\\ &\Leftrightarrow 1+x_k\leq \exp (x_k-{1\over6k}) \end{align}$$ Using $$x+1\leq e^x$$we get $$x_k+1\leq e^{x_k}$$

Answer 1

$log(1+x) - x$ corresponds to the error of a first-order approximation of $log(1+x)$, so this will be least where the second (and higher) derivatives are least, which is on the positive side of 0. This means we can ignore the alternating signs and prove $\frac{1}{\sqrt{k}} - log(1+\frac{1}{\sqrt{k}}) \geq \frac{1}{6k}$.

The series for $log(1+x)$ is $x - \frac{x^2}{2} + \frac{x^3}{3} - ...$, which means we must have $\frac{1}{2k} - \frac{1}{3k^{3/2}} + \frac{1}{4k^2} - ... \geq \frac{1}{6k}$, or $\frac{1}{3k} \geq \frac{1}{3k^{3/2}} - \frac{1}{4k^2} + ...$

But the RHS is less than $\frac{1}{3k^{3/2}}$ for any $k \geq 1$, so the inequality definitely holds in the required range.

Answer 2

Consider the function $f(x) = \log(1+x) - x + \frac 16 x^2$ for $-1 < x < 2$. We have $$ f'(x) = \frac{1}{1+x} -1 + \frac 13 x = \frac{x(x-2)}{3(x+1)} $$ so that $f$ is increasing on $(-1, 0)$ and decreasing on $(0, 2)$. As a consequence, the maximum of $f$ on the interval $(-1, 2)$ is $f(0) = 0$, i.e. $$ \log(1+x) - x \le -\frac 16 x^2 \, . $$ Setting $x = x_k = \frac{(-1)^k}{\sqrt k}$ we get that $$ \log(1+x_k) - x_k \le -\frac{1}{6k} $$ for $k \ge 2$.

Answer 3

I will show that

$$\dfrac{1}{4k(1+1/\sqrt{2k})} \lt x_{2k}-\log(1+x_{2k}) \lt \dfrac{1}{4k} $$ and $$\dfrac{1}{2(2k+1)} \lt x_{2k+1}-\log(1+x_{2k+1}) \lt \dfrac{1}{2(2k+1)(1-1/\sqrt{2k+1})} $$ which implies a stronger result.

$\sum_{k=0}^m t^k =\dfrac{1-t^{m+1}}{1-t} =\dfrac{1}{1-t}-\dfrac{t^{m+1}}{1-t} $ so $\dfrac{1}{1-t} =\sum_{k=0}^m t^k+\dfrac{t^{m+1}}{1-t} $.

Putting $-t$ for $t$, $\frac{1}{1+t} =\sum_{k=0}^m (-t)^k+\frac{(-1)^{m+1}t^{m+1}}{1+t} $.

Integrating from $0$ to $x$,

$\begin{array}\\ \log(1+x) &=\int_0^x \dfrac{dt}{1+t}\\ &=\int_0^x (\sum_{k=0}^m (-t)^k)dt+\int_0^x \dfrac{dt(-1)^{m+1}t^{m+1}}{1+t}\\ &= \sum_{k=0}^m (-1)^k\int_0^xt^kdt+(-1)^{m+1}\int_0^x \dfrac{t^{m+1}dt}{1+t}\\ &= \sum_{k=0}^m (-1)^k\dfrac{x^{k+1}}{k+1}+(-1)^{m+1}\int_0^x \dfrac{t^{m+1}dt}{1+t}\\ &= x-\int_0^x \dfrac{tdt}{1+t} \qquad\text{putting } m=0\\ \end{array} $

If $x > 0$ then

$\begin{array}\\ x-\log(1+x) &=\int_0^x \dfrac{tdt}{1+t}\\ &\lt \int_0^x t\,dt\\ &=\dfrac{x^2}{2}\\ x-\log(1+x) &=\int_0^x \dfrac{tdt}{1+t}\\ &\gt \int_0^x \dfrac{t\,dt}{1+x}\\ &=\dfrac{x^2}{2(1+x)}\\ \end{array} $

so

$\dfrac{x^2}{2(1+x)} \lt x-\log(1+x) \lt \dfrac{x^2}{2} $.

If $x < 0$ then

$\begin{array}\\ x-\log(1+x) &=\int_0^x \dfrac{tdt}{1+t}\\ &=-\int_0^{-x} \dfrac{-tdt}{1-t}\\ &=\int_0^{-x} \dfrac{tdt}{1-t}\\ &<\int_0^{-x} \dfrac{tdt}{1+x}\\ &=\dfrac{(-x)^2}{2(1+x)}\\ &=\dfrac{x^2}{2(1-|x|)}\\ x-\log(1+x) &=\int_0^{-x} \dfrac{tdt}{1-t}\\ &>\int_0^{-x} t\,dt\\ &=\dfrac{x^2}{2}\\ \end{array} $

so

$\dfrac{x^2}{2} \lt x-\log(1+x) \lt \dfrac{x^2}{2(1-|x|)} $.

We have $x_k={(-1)^k\over\sqrt k} $ so $x_{2k}={1\over\sqrt {2k}} $ and $x_{2k+1}=-{1\over\sqrt {2k+1}} $.

Note that $x_k^2 = \dfrac1{k}$ so $\dfrac{x_k^2}{2} = \dfrac1{2k} $.

Therefore,

$\dfrac{1}{4k(1+1/\sqrt{2k})} \lt x_{2k}-\log(1+x_{2k}) \lt \dfrac{1}{4k} $ and $\dfrac{1}{2(2k+1)} \lt x_{2k+1}-\log(1+x_{2k+1}) \lt \dfrac{1}{2(2k+1)(1-1/\sqrt{2k+1})} $.

Therefore $\log(1+x_{2k})-x_{2k} \lt -\dfrac{1}{4k(1+1/\sqrt{2k})} $ and this is less than $-\dfrac1{6(2k)} $ if $12 > 4(1+1/\sqrt{2k})$ or $2 > 1/\sqrt{2k}$ which is always true.

Similarly, $\log(1+x_{2k+1})-x_{2k+1} \lt -\dfrac{1}{2(2k+1)} $ and this is always less than $-\dfrac{1}{6(2k+1)} $.