# Try to prove $\lim_{n \to \infty}n(\ln 2-A_n) = \frac{1}{4}$

$$A_n=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$ Try to prove $$\lim_{n \to \infty}n(\ln 2-A_n) = \frac{1}{4}$$

I try to decompose $\ln 2$ as $$\ln(2n)-\ln(n)=\ln\left(1+\frac{1}{2n-1}\right)+\dots+\ln\left(1+\frac{1}{n}\right)\;,$$ but I can't continue, is that right?

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What exactly is the question? I see a limit, but I think there should be an equality sign somewhere. –  TMM Feb 29 '12 at 14:27
Do you mean that you’re trying to prove that $$\lim_{n \to {\infty}}n(\ln 2-A_n)$$ is equal to $\frac14$? –  Brian M. Scott Feb 29 '12 at 14:27
@BrianM.Scott : Is it possible that the word "prove" was used where the word "evaluate" was needed? We often see confusion about those words and "solve" and the like in this forum. –  Michael Hardy Feb 29 '12 at 14:34
One the other hand, it does appear that the limit is probably $1/4$. –  Michael Hardy Feb 29 '12 at 14:35

## 4 Answers

Trying to follow the idea of the OP, write $$\log 2 = \log (2n+2) - \log (n+1) = \log \left( 1+ \frac{1}{2n+1} \right) + \log \left( 1+ \frac{1}{2n} \right) + \cdots + \log \left( 1+ \frac{1}{n+1} \right)$$

so then $$n( \log 2 - A_n) = n\log \left(1+ \frac{1}{2n+1} \right) + n\sum_{k=1}^n \left( \log \left( 1+ \frac{1}{n+k} \right) - \frac{1}{n+k} \right) .$$

Since near $x=0$ we have $\displaystyle \log(1+x) = x - \frac{x^2}{2} + \mathcal{O}(x^3) ,$ the first term tends to $1/2$ and the summand is $\displaystyle \frac{-1}{2 (n+k)^2 } + \mathcal{O}(1/n^3) .$ Thus, $$n\sum_{k=1}^n \left( \log \left( 1+ \frac{1}{n+k} \right) - \frac{1}{n+k} \right) = \frac{-1}{2} \cdot \frac{1}{n} \left( \sum_{k=1}^n \frac{1}{\left(1+ \frac{k}{n} \right)^2}\right) + \mathcal{O}(1/n)$$

$$\to \frac{-1}{2} \int^1_0 \frac{1}{(1+x)^2} dx= -\frac{1}{4}.$$

Thus, $$n(\log 2 - A_n) \to \frac{1}{2} - \frac{1}{4} = \frac{1}{4}.$$

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Letting $f(x)=1/x$ we have $$\int_1^2 f(x) \; dx = \log_e 2$$ and $$\frac 1 n \left( f\left(1+\frac 1 n\right) + f\left(1+\frac 2 n\right) + f\left(1+\frac 3 n\right) + \cdots + f\left(1+\frac n n\right) \right) \to \int_1^2 f(x) \; dx\text{ as }n\to\infty,$$ so $$\frac 1 n \left( \frac{n}{n+1} + \frac{n}{n+2}+\frac{n}{n+3} + \cdots + \frac{n}{n+n} \right) \to \log_e 2 \text{ as } n\to \infty. \tag{1}$$ Since $f$ is a decreasing function, this is a lower Riemann sum, so it's approaching the integral from below. The difference $$\log_e 2 - \{\text{the sum in (1)}\}$$ is positive and approaches $0$. That difference is the sum of the areas of $n$ regions below the curve and above the tops of the rectangles that you draw when you illustrate the Riemann sum. Each such region is almost a triangle. Its base has length $1/n$. It has a vertical side that is a straight line. Its hypotenuse is a curve that is nearly a straight line. The sum of the heights of those almost-triangles is $1/2$. So the sum of $1/2\times\text{base}\times\text{height}$ is $1/2\times1/n\times1/2$. Multiply it by $n$ to get $1/4$.

But they're not exactly triangles, since the hypotenuse is a curve. It approaches a straight line as $n\to\infty$. The remaining problem is to deal with this present paragraph.

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This sum is also a Riemann lower sum of $\int_n^{2n}s^{-1}ds$ with rectangles of fixed width $1$. In this picture it is quite clear that the error tends to zero since it is dominated by $\sum_{k=n}^{\infty}\varepsilon_k$ where $\varepsilon_k$ is the error contributed by the interval $[k, k+1]$. –  WimC Feb 29 '12 at 15:05

Let's cheat and use one of Euler's many results:

$$\sum_{i=1}^{n} \frac{1}{i} = \ln n + \gamma + \frac{1}{2n} + O\left(\frac{1}{n^2}\right)$$

Note that:

$$A_n + \sum_{i=1}^{n} \frac{1}{i} = \sum_{i=1}^{2n} \frac{1}{i}$$

Substituting Euler's result for both summations, we get:

$$A_n + \ln n + \gamma + \frac{1}{2n} + O\left(\frac{1}{n^2}\right) = \ln 2n + \gamma + \frac{1}{4n} + O\left(\frac{1}{n^2}\right)$$

Rearranging, and using $\ln 2n = \ln 2 + \ln n$, we get

$$A_n = \ln 2 - \frac{1}{4n} + O\left(\frac{1}{n^2}\right)$$

Thus the requested limit becomes

$$\lim_{n \to \infty} n (\ln 2 - A_n) = \lim_{n \to \infty} n \left(\frac{1}{4n} - O\left(\frac{1}{n^2}\right)\right) = \frac{1}{4}$$

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Note that $$\log2-A_n=\int_n^{2n}\frac{\mathrm dx}x-\sum_{k=1}^n\frac1{n+k}=\sum_{k=1}^n\int_0^1\frac{1-x}{(n+k-1+x)(n+k)}\mathrm dx.$$ To get an upper bound, use $n+k-1+x\geqslant n+k-1$, hence $$\log2-A_n\leqslant\sum_{k=1}^n\frac{1}{(n+k-1)(n+k)}\int_0^1(1-x)\mathrm dx.$$ The integral is $\frac12$ and the sum is $$\sum_{k=1}^n\left(\frac{1}{n+k-1}-\frac1{n+k}\right)=\frac1n-\frac1{2n}=\frac1{2n},$$ hence $$\log2-A_n\leqslant\frac1{4n}.$$ To get a lower bound, use $n+k-1+x\leqslant n+k\leqslant n+k+1$, hence $$\log2-A_n\geqslant\sum_{k=1}^n\frac{1}{(n+k)(n+k+1)}\int_0^1(1-x)\mathrm dx.$$ The integral is still $\frac12$ and the sum is $$\sum_{k=1}^n\left(\frac{1}{n+k}-\frac1{n+k+1}\right)=\frac1{n+1}-\frac1{2n+1}=\frac{n}{(n+1)(2n+1)},$$ hence $$\log2-A_n\geqslant\frac{n}{2(n+1)(2n+1)}=\frac{u_n}{4n},$$ with $$u_n=\frac1{(1+1/n)(1+1/(2n))}.$$ Finally, $\frac14u_n\leqslant n(\log2-A_n)\leqslant\frac14$ and $u_n\to1$ hence $n(\log2-A_n)\to\tfrac14$.

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