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

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 $$x_k+1\leq e^{x_k}$$

$$log(1+x) - x$$ 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.

• Thank you. Why does ${1\over 3k^{3/2}}-{1\over 4k^2}+...<{1\over 3k^{3/2}}$? – J. Doe Jun 17 at 18:52
• If you subtract $\frac{1}{3k^{3/2}}$ from both sides, you can place the remaining terms into pairs containing a negative and positive term. As $k$ increases, the positive term in each pair decreases faster than the corresponding negative term because it has a larger exponent in the denominator, so all that remains is showing that the sum is negative for $k=1$, which is trivial. – AxiomaticSystem Jun 17 at 18:56

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 $$k \ge 2$$.

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 $$-\dfrac{1}{6(2k+1)}$$.