Let a series be given by $$\sum_{k=1}^{\infty}\frac{\left \lceil \frac{k}{\pi} \right \rceil}{k^4}$$Show that it converges and that the convergent value is in $[1,11/10]$
This is a strange one as I am not sure on how to to show the second part of the question. The first one - according to my attempt - would just be to show convergence by comparison. I.e just take the $p$-sreies with $p=2$ (that converges) and for every $n$ in the sequences the $p$-series is at least equal to the one in question or greater and thus it must converge by comparison.
How would I from analytical theory argue that the value (whatever the value is) is in that interval?
Since the $k=1$ summand contributes $\frac{\lceil 1/\pi \rceil}{1^4}=1$, we need to prove that $0\le \sum_{k=2}^{\infty}\frac{\left \lceil k/\pi \right \rceil}{k^4}\le\frac1{10}$. The first inequality is obvious (in other words, $k=1$ already gives the "sufficiently long finite prefix" mentioned in Troposphere's comment). As for the upper bound, we write $\sum_{k=2}^{\infty}\frac{\left \lceil k/\pi \right \rceil}{k^4} = \frac1{16}+\frac1{81}+\frac2{256}+\sum_{k=5}^{\infty}\frac{\left \lceil k/\pi \right \rceil}{k^4}$, and estimate \begin{align*} \sum_{k=5}^{\infty}\frac{\left \lceil k/\pi \right \rceil}{k^4} &< \sum_{k=5}^{\infty}\frac{(k/\pi+1)}{k^4} \\ &= \frac1\pi \sum_{k=5}^{\infty}\frac1{k^3} + \sum_{k=5}^{\infty}\frac1{k^4} \\ &< \frac1\pi \int_4^\infty \frac1{x^3}\,dx + \int_4^\infty \frac1{x^4}\,dx \\ &= \frac1\pi \frac{-1}{2x^2}\bigg|_4^\infty + \frac{-1}{3x^3}\bigg|_4^\infty = \frac1{32\pi}+\frac1{192}, \end{align*} where the latter inequality is valid because the sums are right-hand Riemann sums for these integrals of decreasing functions (drawing the picture reminds us of the inequality).
Therefore $\sum_{k=2}^{\infty}\frac{\left \lceil k/\pi \right \rceil}{k^4} < \frac1{16}+\frac1{81}+\frac2{256}+\frac1{32\pi}+\frac1{192} < 0.098 < \frac1{10}$, as needed. (How did I know to single out the $k=2,3,4$ terms? I didn't!—I just tried bounding the infinite sum, saw three times in a row that the estimate wasn't quite good enough, and thrice singled out another exact term until the whole thing worked.)