convergence help

**friday616** · November 7th 2009, 02:24 PM

For each positive integer n, let

{y_n}= 1 + 1/2 +...+1/n- the integral of (1/x) from 1 to n.

Prove that the sequence {y_n} converges.

**redsoxfan325** · November 7th 2009, 06:23 PM

Originally Posted by friday616

For each positive integer n, let

{y_n}= 1 + 1/2 +...+1/n- the integral of (1/x) from 1 to n.

Prove that the sequence {y_n} converges.

Proof of monotone decreasing: $y_{n-1}-y_n=\sum_{k=1}^{n-1}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}+\int_1^n\frac{1}{x}\,dx-\int_1^{n-1}\frac{1}{x}\,dx$ $=\int_{n-1}^n\frac{1}{x}\,dx-\frac{1}{n}\geq\int_{n-1}^n\frac{1}{n}\,dx-\frac{1}{n}=0$

To prove that $\{y_n\}$ is bounded below by $0$ , if $P$ is the partition $\{x_1=1,x_2=2,...,x_n=n\}$ , and $f(x)=\frac{1}{x}$ , the upper sum $U(P,f)$ of the integral of $f(x)$ is $\sum_{k=1}^{n-1}\frac{1}{x_k}(x_{k+1}-x_k)=\sum_{k=1}^{n-1}\frac{1}{k}$ . Since $\sum_{k=1}^n\frac{1}{k}>U(P,f)>\int_1^n\frac{1}{x} \,dx$ , it follows that $y_n>0$ for all $n$ .

Since $\{y_n\}$ is a monotonically decreasing sequence bounded below, it converges.

This is actually a famous limit and converges to the Euler-Mascheroni Constant, $\gamma$ .

$\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k}-\ln(n)\right)=\gamma\approx0.577$

You're welcome.

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