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Here is a solution: $$ \lim_{n\to \infty}\int_1^n\frac{1}{x^p}\,dx=\boxed{\lim_{n\to\infty}\left[\frac{1}{1-p}\frac{1}{x^{p-1}}\right]_1^n}=\lim_{n\to\infty}\frac{1}{1-p}\left(\frac{1}{n^{p-1}}-1\right)=\frac{1}{p-1} $$ if $p>1 \Rightarrow$ converges.

Solution is correct.

Kindly explain the first step, which is enclosed in a box.

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You evaluate the integral first. Assuming $p\ne1$, an antiderivative of ${1\over x^p}=x^{-p}$ is ${1\over -p+1}x^{-p+1}={1\over1-p}{1\over x^{p-1}}$. – David Mitra Feb 5 '13 at 13:59

The first step is saying that an antiderivate of the function $\displaystyle\frac{1}{x^p}$ is $\displaystyle\frac{1}{1-p}\frac{1}{x^{p-1}}$ which you can check by verifying $$\frac{d}{dx}\left(\frac{1}{1-p}\frac{1}{x^{p-1}}\right) = \frac{1}{x^p}.$$

If $F$ is an antiderivative of $f$ (that is $F' = f$), then (one half of) the Fundamental Theorem of Calculus says that $\int_a^bf(x)dx = [F(x)]_a^b = F(b) - F(a)$.

Added Later: As David Mitra points out, the first paragraph only applies when $p \neq 1$.

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