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Is $\lim\limits_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} = 0$?

Show that

$$\ln(2) = \lim_{n\rightarrow\infty}\left( \frac{1}{n + 1} + \frac{1}{n + 2} + ... + \frac{1}{2n}\right)$$

by considering the lower Riemann sum of $f$ where $f(x) = \frac{1}{x}$ over $[1, 2]$

I was confused looking at the equality to begin with, since taking $n \rightarrow \infty$ for all of those terms would become $0$ right?

Anyway, I attempted it regardless.

$$\sum_{k=1}^n \frac{1}{n}(f(1 + \frac{k}{n}))$$

$$= \sum_{k=1}^n \frac{1}{n}(\frac{1}{1+ \frac{k}{n}})$$

$$=\sum_{k=1}^n \frac{1}{n + k} = $$ the sum from the question?

I wasn't sure what to do from here. I tried something else though:

$$=\frac{1}{n + \frac{n(n+1)}{2}}$$

$$=\frac{2}{n^2 + 3n}$$ which seemed equally useless if I'm taking $n \rightarrow \infty$ as it all becomes $0$.

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Your "something else" is completely invalid. It looks like you tried to move the fraction symbol outside of the sum symbol somehow and this just isn't possible. In any case, it is true that the individual terms tend to zero as $n \to \infty$, but it is also true that the number of terms tends to infinity, so you can't conclude that their sum is zero. For example, the sum of $n$ copies of $\frac{1}{n}$ is $1$ and the sum of $n$ copies of $\frac{1}{2n}$ is $\frac{1}{2}$, so your sum is between $\frac{1}{2}$ and $1$. –  Qiaochu Yuan Jun 7 '12 at 15:55
    
Your derivation of the sum is correct, it might be nice to point out which intervals you are using, your choice of endpoint (left/right/...) and that 1/n is the width of each; and then of course take the limit. Note that (when the details are added) that you have done all that was asked--show that ln 2 equals the limit of the sum, te problems asks for no more. –  Chris K. Caldwell Jun 7 '12 at 16:08
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marked as duplicate by Marvis, Jyrki Lahtonen, Hans Lundmark, Qiaochu Yuan Jun 7 '12 at 17:47

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3 Answers

up vote 6 down vote accepted

Although the terms of the sum go to $0$ as $n\to0$, there are $n$ terms and each is between $\dfrac1n$ and $\dfrac{1}{2n}$, therefore, the sum should be between $1$ and $\dfrac12$.


There at least a couple of ways to look at the sum $\displaystyle\sum_{k=1}^n\frac{1}{n+k}$:

Bound by Integrals:

Since $\frac1x$ is a decreasing function, we get that

$$ \log\left(\frac{2n+1}{n+1}\right)=\int_{1}^{n+1}\frac{\mathrm{d}x}{n+x}\le\sum_{k=1}^{n}\frac{1}{n+k}\le\int_{0}^{n}\frac{\mathrm{d}x}{n+x}=\log(2)\tag{1} $$ Both sides of $(1)$ go to $\log(2)$ as $n\to\infty$, and the sum is sandwiched in between.

Riemann Sums:

The sum can be rewritten as $$ \sum_{k=1}^n\frac{1}{1+\frac{k}{n}}\frac1n\tag{2} $$ which is a Riemann sum for the integral $$ \int_0^1\frac{1}{1+x}\mathrm{d}x\tag{3} $$ where $\frac{k}{n}$ approximates $x$ and $\frac1n$ approximates $\mathrm{d}x$.

As $n\to\infty$, the mesh represented in $(2)$ gets finer and the sum approaches the integral, whose value is $\log(2)$.


Note that $\log(2)\,\dot{=}\,0.69314718$ is between $1$ and $\frac12$, as mentioned above.

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When converting the Riemann integral to a 'normal' integral, how do you determine the limits of integration? Is it just always from $0$ to 'the size of the segment'? How do you know generally for any given Riemann integral? –  stariz77 Jun 7 '12 at 17:10
    
@stariz77: I assume you are asking about converting the Riemann sum to a Riemann integral. In the case above, I mentioned that $\frac{k}{n}$ approximates $x$, this means that since $k$ ranges from $1$ to $n$, $\frac{k}{n}$ ranges range from $0$ to $1$ and so would $x$. In general, if the Riemann sum is written as $$\sum f\left(\frac{k}{n}\right)\frac{1}{n}$$ the limits of $x$ would be the range of $\frac{k}{n}$. –  robjohn Jun 7 '12 at 17:16
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You were almost there!$$\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n+k}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{n}\frac{1}{1+\frac{k}{n}}=\int_0^1\frac{dx}{1+x}=\ln 2$$

The central equality above follows from the fact we know the integral on the right exists and thus it equals the limit of its Riemann sums no matter how we make those sums in the limit (as long, of course, as the interval's partition mesh tends to zero)

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I guess there's a typo in your last line. "mesh tends to zero" –  Gigili Jun 7 '12 at 16:07
    
No, but I'm afraid this could be a language problem. What I meant was: in the Riemann integral, the limit of the Riemann sums must exists finitely independently of what points are taken within each subpartition interval (for $f(c_i)$) and how the partitions are taken as long as the partitions' mesh approaches zero, and here "a partition's mesh" is the maximum of the lengths of the partition's intervals. I hope this clarifies things, but if you think I'm misusing some term please do write back. –  DonAntonio Jun 7 '12 at 16:16
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Okay, you're right. I'm not sure why I thought there's a typo now that I read more carefully. thanks for clarification and plus one. –  Gigili Jun 7 '12 at 16:58
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METHOD I

We may recall the celebre limit that yields Euler-Mascheroni constant, namely:

$$\lim_{n\to\infty} 1+\frac1{2}+\cdots+\frac{1}{n}-\ln{n}={\gamma}$$ $\tag{$\gamma$ is Euler-Mascheroni constant}$ Then everything boils down to: $$\lim_{n\to\infty}\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} = \lim_{n\to\infty}{\gamma}+\ln{2n}-{\gamma}-\ln{n}= \ln{2}.$$

METHOD II

Use one of the consequences of the Lagrange's theorem applied on $\ln(x)$ function, namely:

$$\frac{1}{k+1} < \ln(k+1)-\ln(k)<\frac{1}{k} \space , \space k\in\mathbb{N} ,\space k>0$$

Taking $k=n,n+1,...,2n$ values to the inequality and then summing all relations, we get all we need in order to apply Squeeze theorem.

METHOD III

We may use Botez-Catalan identity and immediately get that:

$$\lim_{n\to\infty}\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} = \lim_{n\to\infty} 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + (-1)^{2n+1}\frac{1}{2n}= $$ $$\lim_{n\to\infty} 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + (-1)^{n+1}\frac{1}{n}=\ln{2}.$$ The last series' limit is obtained by using Taylor expansion of $\ln(x+1)$ and take $x=1$

The proofs are complete.

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I like seeing so many different approaches to the same solution! (+1) I have used the Botez-Catalan (not knowing it had that name) in developing an algorithm to compute $\gamma$ with which I computed 10000 digits. –  robjohn Jun 7 '12 at 20:12
    
@robjohn: that's great, and looks nice your work. I'm there right now. :-) –  Chris's sis Jun 7 '12 at 20:16
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