Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$A_n=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$ 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?

share|improve this question
    
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 4

up vote 2 down vote accepted

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

share|improve this answer

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.

share|improve this answer
    
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

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

share|improve this answer

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 $n(\log2-A_n)\to\tfrac14$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.