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Is $a_n = \frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}$ convergent? If so find its limit.

Heres what ive tried:

$... < \frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n} < \frac{n}{n} = 1$

What should I put on the left side of the inequality? I tried $\frac{n}{2n}$ but that tends to a half and I cant think of solving this in any other way except the squeeze rule.

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marked as duplicate by Daniel Fischer real-analysis Jan 2 '18 at 19:02

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    $\begingroup$ You could try integrating $f(x)=\frac 1x$ between appropriate limits to get a tighter comparison - suggest drawing a diagram. $\endgroup$ – Mark Bennet Jan 2 '18 at 11:15
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You can write that for $n \in \mathbb{N}^{*}$ $$ a_n=\sum_{k=n}^{2n}\frac{1}{k} $$ What I use to proof this ( without using an already known equivalence ) is to say that the function $\displaystyle x \mapsto \frac{1}{x}$ is decreasing ( and is positive and tends to $0$ ). Hence if you draw a decreasing function you can bound the function on $[n,n+1]$ by two rectangular area : a small and a big one. As shown below :

So it is written as $$ \int_{k}^{k+1}\frac{\text{d}t}{t} \leq \frac{1}{k} \leq \int_{k-1}^{k}\frac{\text{d}t}{t}$$

By summing with Chasles relation with $n \geq 2$ $$ \int_{n}^{2n+1}\frac{\text{d}t}{t} \leq a_n \leq \int_{n-1}^{2n}\frac{\text{d}t}{t}$$ then $$ \ln\left(\frac{2n+1}{n}\right) \leq a_n \leq \ln\left(\frac{2n}{n-1}\right)$$

Then the sequence $\displaystyle \left(a_n\right)_{n \in \mathbb{N}^{*}}$ converges and $\displaystyle a_n \underset{n \rightarrow +\infty}{\rightarrow}\ln\left(2\right)$.

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This is clearly a Riemann sum for $k=0$ to $k=n$ as $n \to\infty$. So,

$$a_n=\lim_{n \to \infty}\frac{1}{n}\sum_{k=0}^{n}\frac{1}{1+\frac{k}{n}}$$

$$\int_0 ^1 \frac{1}{1+x}dx=\ln2$$

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Rewrite the sum as $\frac{1}{n}+\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}$.

The first term $\frac{1}{n}\to 0$ when $n\to\infty$.

As for the second term, notice that it is a (lower) Riemann sum for $\int_0^1\frac{dx}{1+x}$ so, when $n\to\infty$, its limit exists and is equal to $\int_0^1\frac{dx}{1+x}=\log 2-\log 1=\log 2$.

Finally, this means that the limit of the original sequence is $\log 2$.

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  • $\begingroup$ @Teepeem You are right, thanks for the correction. I forgot to type this factor $\frac{1}{n}$. I still need to treat either the term $\frac{1}{n}$ or $\frac{1}{2n}$ separately because the OP's sum has $n+1$ terms - one too much for the Riemann sum argument. Please see my edited answer. $\endgroup$ – user491874 Jan 2 '18 at 18:47
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It is same as $\lim_{n\to\infty}a_n=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{\frac{1}{n}}{1+\frac{k}{n}}=\int_{0}^{1}\frac{1}{1+x}dx$.

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The Euler-Maclaurin summation formula captures the intuition that we can approximate the sum by an integral. It gives: $$\sum_{y<k\leq x}f(k) = \int_y^xf(t)dt+\int_y^x\{t\}f'(t)dt-\{x\}f(x)+\{y\}f(y)$$ for $f\in C^1[y,x]$ continuously differentiable, where $\{\cdot\}$ is the fractional part.

In particular, when $f\to0$ at $\infty$ and $\int_1^\infty f'<\infty$, this gives $\sum_{y\leq k\leq x}f(k) = \int_y^xf(t)dt+\varepsilon(x,y)$ where $\varepsilon\to0$ for $y\to\infty$.

In this case, with $y=n, x=2n, f(t)=1/t$:

$$\sum_{n\leq k\leq 2n}\frac1k = \int_n^{2n}\frac{dt}t+o(1) = \log 2+o(1)$$ (where $o(1)$ is little o notation, meaning that it goes to $0$ as $n\to\infty$). One can make the difference explicit and prove that it is at most $2/n$.

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