Estimating sums by integrals Estimating sums by integrals. Let $f : \mathbb{N}→\mathbb{N}$ be an increasing function. Show that $$\sum \limits_{i=1}^n \frac1{f(i)}<\frac1{f(1)}+\int \limits_{1}^{n}\frac1{f(x)}dx$$
I really need help on this. There is barely anything to refer back to on my lecture notes.
All I know is that the function seems to be continuous because zero is not a natural number so there is no divisors with zero.
I don't understand how the summation can find the area because all it finds is the image values and sums them up. That's not the same as area is it.

We first show that 1/f(n+1) < Integral(from x = n to x = n+1)(1/f(x))dx, for all n.
Let x be in [n,n+1]:
Then, since f is increasing, f(x) <= f(n+1), which gives: 1/f(n+1) <= 1/f(x), thus:
Integral(from x = n to x = n+1) (1/f(n+1))dx <= Integral(from x = n to x = n+1) (1/f(x))dx
Thus: 1/f(n+1) <= Integral(x = n to x = n+1) (1/f(x))dx.
Now, argue by induction on n:
For n = 1, Sum(i=1 to 1) (1/f(i)) = 1/f(1) <= 1/f(1) + etc.; so the property holds.
Assume this holds up to n. Now:
Sum(i=1 to n+1) (1/f(i)) = Sum(i=1 to n) (1/f(i)) + 1/f(n+1)
< 1/f(1) + Integral(x = 1 to x = n) (1/f(x))dx + 1/f(n+1) (By induction hypothesis).
< 1/f(1) + Integral(x = 1 to x = n) (1/f(x))dx + integral(x = n to x = n+1) (1/f(x))dx (By what has been shown above)
= 1/f(1) + Integral(x = 1 to x = n+1) (1/f(x))dx (By Chasles's relation).
Therefore the property holds by induction on n, for any natural number.
