old qualifying problem Suppose that ${\{c_n\}}_{n=1}^{\infty}$ is a sequence of nonnegative numbers.
(a) If $\sum_{n=1}^{\infty} c_n <\infty$, show that there is a C such that $\\$
(1)    #$\{ n: c_n>s\}\leq \frac{C}{s}$ for all $s>0$ 
(above, # denotes cardinality)
(b) Give an example of a sequence ${\{c_n\}}_{n=1}^{\infty}$ which  satisfies (1), but $\sum_{n=1}^{\infty} c_n =\infty$
(c) If, however, #$\{ n: c_n>s\}\leq \frac{C}{\sqrt{s}}$ for s>0, show that $\sum_{n=1}^{\infty} c_n <\infty$,
I am not sure about my solution(probably wrong),
for (a) since $\sum_{n=1}^{\infty} c_n <\infty$ then $c_n \rightarrow 0$ then there is a N>0, when n>N we have $c_n<s$, so #$\{ n: c_n>s\} = N$, so we have a $C=Ns$
for (b) let $c_n = s$ for all n, so #$\{ n: c_n>s\} = 0 \leq \frac{C}{s}$ and satisfy $\sum_{n=1}^{\infty} c_n =\infty$
for (c) I'm not sure about this, if we let s=1 and $c_n = 1$, then #$\{ n: c_n>s\} = 0$ and $\sum_{n=1}^{\infty} c_n =\infty$, so I am confused. What's the meaning of cardinality?
 A: For (a) your proof is not correct since the $C$ you chose depends on $s$. Note that the assumption in (a) is exactly the statement that $c:\mathbb{N}\to \mathbb{R}$ is in $L^1$ (counting measure). Thus by Chebyshev's inequality we have $\#\{n: c_n > s\} \leq \frac{1}{s} ||c||_1$. Since $||c||_1 = \sum c_n$ we find that the choice $C:= \sum c_n$ suffices.
For (b) your proof is also wrong because your function $c$ depends on $s$ and not just $n$. Instead take $c_n = 1/n$. Then obviously $\sum c_n = \infty$ but $\#\{ n : 1/n > s\} = \lfloor \frac{1}{s} \rfloor \leq \frac{1}{s}$.
For (c) the question means for all $s>0$, not just for some $s>0$. Then 
$$
\sum_{n=0}^\infty c_n = \sum_{n=0}^\infty \int_0^\infty 1_{t < c_n} dt
=\int_0^\infty \sum_{n=0}^\infty  1_{t < c_n} dt =\int_0^1 \sum_{n=0}^\infty  1_{t < c_n} dt + \int_1^\infty \sum_{n=0}^\infty  1_{t < c_n} dt
$$
$$
\leq \int_0^1 \frac{C}{\sqrt{t}} dt+ \sum_{n=0}^\infty \int_1^\infty 1_{t < c_n} dt
$$
$$
=\int_0^1 \frac{C}{\sqrt{t}} dt+ \sum_{n=0}^\infty \max(c_n-1,0)
$$
The left integral is finite, and the right sum is actually a finite sum since there can only be finitely many $c_n > 1$ since $\#\{n: c_n > 1\} \leq C/1$. Thus $\sum c_n < \infty$.
A: As @nulluser points out, your solution to (a) can't be right, and the series $\frac{1}{\sqrt{n}}$ is an example of a series whose terms tend to zero, but which satisfies the condition (so you need to use something more than terms tending to zero).
For part (c), I recommend using the fact that you can rearrange terms in a non-negative series and get the same result. Try grouping the terms by the given condition (first count all the terms bigger than $1/2$, then all the terms between $1/3$ and $1/2$, and so forth) and see if you can show convergence that way.
