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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.

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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.

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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 $$\;f(n+1)\le\int_n^{n+1}f(t)dt\le f(n)$$

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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"

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