When the Integral Converges
Upper Estimates
We have
$$
f(2) \le \int_1^2 f(x) \,dx,
$$
and, similarly,
$$
f(3) \le \int_2^3 f(x)\, dx.
$$
In general,
$$
f(n) \le \int_{n-1}^{n} f(x) \,dx.
$$
Therefore, we can conclude that
$$
s_n -f(1) \le \int_1^n f(x) \,dx,
$$
which gives us
$$
s_n \le f(1) + \int_1^n f(x)\, dx.
$$
Taking the limit as $n\to\infty$ yields
$$
\sum_{n=1}^\infty f(n) \le\, f(1) + \int_1^\infty f(x)\, dx$$
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Converging Integral
If the integral converges, then we can conclude that $\sum f(n)$ must also converge, as, if not, we would have found a number bigger than $\infty$!
Tail Estimates
Putting together the lower estimate from the previous page and the above upper estimate yields:
Tail Estimates:
$$
\int_1^\infty f(x) \,dx \le \sum_{n=1}^\infty f(n) \le\, f(1) + \int_1^\infty f(x)\, dx$$
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