Integral Test For Convergence - Proof

Proof

The proof basically uses the comparison test, comparing the term f(n) with the integral of f over the intervals and, respectively.

Since f is a monotone decreasing function, we know that

f(x)\le f(n)\quad\text{for }x\in[n,\infty)

and

f(n)\le f(x)\quad\text{for }x\in,

hence for every n larger than N

\int_n^{n+1} f(x)\,dx \le\int_{n}^{n+1} f(n)\,dx =f(n) =\int_{n-1}^{n} f(n)\,dx \le\int_{n-1}^n f(x)\,dx.

Since the lower estimate is also valid for f(N), we get by summation over all n from N to some larger integer M

\int_N^{M+1}f(x)\,dx\le\sum_{n=N}^Mf(n)\le f(N)+\int_N^M f(x)\,dx.

Regarding the last two terms of the inequality: for every (in the summed series), a (in the integral term) is greater than the former term. Working backward from (the final term in the summation), there is an integral increment that is greater than its summation term, except for (because the domain is bounded by N), which is then added to the integral side of the inequality. Thus,

Letting M tend to infinity, the result follows.

Read more about this topic:  Integral Test For Convergence

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