# Proving a limit exists under some conditions

Suppose that $$f(x) > 0$$ is integrable and monotone decreasing on $$[0, \infty)$$. Let $$F_{n} = \int_{0}^{n} f(t) \mathop{dt}$$, $$n = 1, 2, 3, \ldots$$. Prove that

$$\lim_{n\to\infty} F_{n}$$

exists if and only if $$\sum_{n = 1}^{\infty} f(n) < \infty$$.

Hint: Consider $$F(n + 1) - F(n)$$

I'm not sure about how to approach this problem. This is a practice problem that I have for a final exam coming soon. I think that the hint helps us show that the sequence is monotone decreasing because if we can show that quantity is less than $$0$$, it would imply that the terms are getting smaller and smaller.

Hint

Note that $$\sum_{j=1}^n f(j)\leq \int_0^n f(t)\, dt=\sum_{j=1}^n\int_{j-1}^{j} f(t) \, dt\le \sum_{j=1}^n f(j-1)$$ where we used linearity of the integral for the middle equality and the fact that $$f$$ is monotone decreasing for the outer inequalities.

hint

For $$n\ge 0,$$

$$F(n+1)-F(n)=\int_n^{n+1}f(t)dt$$

$$(\forall t\in[n,n+1])\;$$ $$f(n+1)\le f(t)\le f(n)$$ and $$\;f(n+1)\le\int_n^{n+1}f(t)dt\le f(n)$$

I am also studying for my real analysis exam, so if this proof is wrong please let me know so I know what to look over :)

If $$\sum_{n=1}^\infty f(n) < \infty$$ $$\implies$$ $$\lim_{n\rightarrow \infty} F_n < \infty$$

Since $$f$$ is monotonically decreasing on $$[0,\infty)$$, we know that $$f(n+1) < f(n)$$ $$\forall n \in \mathbb{N}$$, and we know $$f(0)$$ exists.

From this we can see that the following comparison is true $$F_n = \int_0^n f(t)dt < \sum_{k=0}^n f(k) = f(0) + \sum_{k=1}^n f(k)$$ now we can take the limit of both sides and see that $$\lim_{n\rightarrow \infty} F_n < f(0) + \sum_{k=1}^\infty f(k)$$ Therofore, the limit must exist because we know that the sum exists, and $$f(0)$$ exists.

To prove the other direction the method is the same, except we say that

$$\sum_{k=1}^n f(k) < \int_0^n f(t)dt$$

since it is not a "right-Riemann Sum"