This function seems to defy the integral test, where am I going wrong? I recently was working on a question posted in an AP calculus BC multiple choice sheet which asked:
Let f(x) be a positive, continuous deceasing function. If $\int_1^∞ f(x)dx$ = 5, then which of the following statements must be true about the series $\sum_1^∞f(n)$?
(a) $\sum_1^∞f(n)$ = 0
(b) $\sum_1^∞f(n)$ converges, and $\sum_1^∞f(n)$ < 5
(c) $\sum_1^∞f(n)$ = 5
(d) $\sum_1^∞f(n)$ converges, and $\sum_1^∞f(n)$ > 5
(e) $\sum_1^∞f(n)$ diverges.
I assumed, due to the integral test, the answer would be (b). However, the answer sheet claimed (d). I thought that this was merely a mistake, however I tried to find a function that met these conditions ( f(x) is strictly positive, f '(x) is negative, f(x) is continuous $\forall$x $\in$ [1,∞) and $\int_1^∞ f(x)dx$ = 5) and seemed to have found one (where the conditions were confirmed via online sources) and (d) seems to be true, thus I'm asking where this function fails to meet the above conditions or how (d) can be true given the integral test contradicting it (from my understanding).
If F(x) = $\int f(x)dx$, then $\lim_{x\to ∞}$F(x) - F(1) must equal five.
Let F(x) = (x - 1)$\biggr(\frac{5}{x+2}-\frac{1}{(x+4)^2}\biggr)$
Then f(x), which we'll let equal $\dfrac{d}{dx}$F(x), equals $\frac{5}{(x+2)}$ - $\frac{1}{(x+4)^2}$ + (x - 1)$\biggr(\frac{-5}{(x+2)^2} +\frac{2}{(x+4)^3}\biggr)$. By plugging f(x) and its derivative into an online graphing calculator, I find that f(x) is strictly positive and its derivative strictly negative. Thus $\sum_1^∞f(n)$ < $\int_1^∞f(x)dx$, but when I use an online source to find these values I find the opposite result:
f(x) in a graphing calculator (Desmos) (note only from 1 to a large number)
https://www.desmos.com/calculator/kfeqcphona
f'(x) in a graphing calculator (Desmos) (I just put f'(x) in a form where f(x) is derived term by term)
https://www.desmos.com/calculator/fnroc3w9rm
$\sum_1^∞f(n)$ = $\sum_1^∞\frac{5}{n+2}-\frac{1}{(n+4)^2}+(n-1)(\frac{-5}{(n+2)^2}+\frac{2}{(n+4)^3})$ = −10ζ(3)−$\frac{7255}{864}$+8(π)23≈5.90138529723494 
According to https://www.emathhelp.net/calculators/calculus-2/series-calculator/.
Where am I going wrong?
 A: Since:


*

*$\displaystyle\int_1^2f(x)\,\mathrm dx<f(1)$

*$\displaystyle\int_2^3f(x)\,\mathrm dx<f(2)$

*$\displaystyle\int_3^4f(x)\,\mathrm dx<f(3)$


and so on, you have$$\int_1^{+\infty}f(x)\,\mathrm dx<\sum_{n=1}^\infty f(n).$$
A: Note that since the function is decreasing


*

*$f(1)>\int_1^2 f(x)dx, \quad f(2)>\int_2^3 f(x)dx, ... \implies   
   \sum_1^\infty f(n)> \int_1^\infty f(x)dx=5$


and the series converges since


*

*$f(2)<\int_1^2 f(x)dx, \quad f(3)<\int_2^3 f(x)dx, ... \implies
   \sum_2^\infty f(n)< \int_1^\infty f(x)dx \\\implies
   \sum_1^\infty f(n)< f(1)+\int_1^\infty f(x)dx$

A: Observe that
$$
\int_{1}^nf(x)\,dx=\sum_{i=1}^{n-1}\int_{i}^{i+1}f(x)\,dx<\sum_{i=1}^{n-1}\tag{1}
f(i)$$
for any integer $n\ge 1$ where we have using the fact that $f$ is decreasing for the inequality. Let $n\to\infty$ in (1) to get that
$$
5\leq\sum_{i=1}^\infty f(i)
$$
A: As the other answers point out quite well, because your function is positive and decreasing the sum will be larger than the integral. To get at your question though: no, this does not defy the integral test. The integral test does not give an upper bound on the sum, but merely says if one (either the sum or integral) converges then so will the other (either the integral or the sum), which is the case here. We need more information to compare the sizes of the two. 
A: For visualization purposes, explore $f(x) = (1/x) \Rightarrow \int_{1}^n f(x)dx = ln(n).$ 
$\sum_{i=1}^{n-1} (1/i) > ln(n) > \sum_{i=2}^n (1/i).$
