Show that there is a $C$ such that $\frac{1}{n^{1+\alpha}} \leq C(\frac {1}{n^\alpha}-\frac{1}{(n+1)^\alpha})$ 
1.(a) Show that if $\alpha>0$, then there is a constant $C$ such that for any $n\in \mathbb N$,
$$\frac{1}{n^{1+\alpha}} \leq C(\frac {1}{n^\alpha}-\frac{1}{(n+1)^\alpha})$$
[Suggestion: Write
$$\frac {1}{n^\alpha}-\frac{1}{(n+1)^\alpha}=\frac{1}{n^\alpha}[\frac{1-\frac{1}{(1+\frac{1}{n})^\alpha}}{\frac{1}{n}}],$$
and estimate the expression in square brackets as converging to a derivative as $n\to\infty$]

In this equation, I am wondering if I set $C$ to be infinity then I can confirm that there is surely a $C$ that exist? Or if not, can anyone tell me how should I interpret this $C$?
 A: define a function $f:[1-\delta, +\infty)\to \mathbb{R}$ (for some small $\delta>0$) as $f(x)=\displaystyle\frac {1}{x^{\alpha}}$ for some $\alpha>0$. Check for yourself that $f$ satisfies all the required conditions for mean value theorem.
so there is a point $c\in [n,n+1]$ satisfying 
$f(n+1)-f(n)=f'(c)= -\displaystyle\frac{1}{\alpha c^{1+\alpha}}$
consequently, $f(n)-f(n+1)=\displaystyle\frac{1}{\alpha c^{1+\alpha}}\geq\displaystyle\frac{1}{\alpha (1+n)^{1+\alpha}}\geq (\displaystyle\frac{1}{\alpha n^{1+\alpha}}).(\displaystyle\frac{n}{n+1})^{1+\alpha}\geq (\displaystyle\frac{1}{\alpha n^{1+\alpha}}).(\displaystyle\frac{1}{2})^{1+\alpha}$
So, $C^{-1}=\displaystyle\frac{1}{\alpha.2^{1+\alpha}}$ works $\cdots$
A: We choose $\displaystyle C=1+\dfrac{1}{\alpha}$ if $\alpha\ge 1$ and $C=1+\dfrac{2}{\alpha(\alpha+1)}$ otherwise. The inequality $\displaystyle \dfrac{1}{n^{1+\alpha}}\le C\left(\dfrac{1}{n^\alpha}-\dfrac{1}{(n+1)^\alpha}\right)$
is equivalent to $$\left(1+\dfrac{1}{n}\right)^\alpha\ge\dfrac{C}{C-\frac{1}{n}}=1+\dfrac{1}{Cn-1}$$
and the left-handed side above, when $\alpha\ge 1$, is $\ge 1+\dfrac{\alpha}{n}$, thus, it suffices to show that $$1+\dfrac{\alpha}{n}\ge 1+\dfrac{1}{Cn-1}\Longleftrightarrow n\left(C-\dfrac{1}{\alpha}\right)\ge 1$$ which is true for $C$ we had chosen. 
For the case $0< \alpha <1$ the left-handed side of the above inequality, according to Taylor expansion, is $1+\dfrac{\alpha}{n}+\dfrac{\alpha(\alpha-1)}{2}\cdot\dfrac{1}{n^2}+\left(\left(\dfrac{\alpha(\alpha-1)(\alpha-2)}{3!}\cdot\dfrac{1}{n^3}\right)\left(\dfrac{4n+(\alpha-3)}{4n}\right)\dots\right)$ which in the last bracket is greater than or equal to zero each term, for $\alpha$ in this range.
We suffice to prove that $\displaystyle \dfrac{\alpha}{n}\left(1-\dfrac{1-\alpha}{2n}\right)=\dfrac{\alpha}{n}\left(1+\dfrac{\alpha-1}{2n}\right)\ge \dfrac{1}{Cn-1}$ for the chosen $C$
we have $\displaystyle 1-\dfrac{1-\alpha}{2n}\ge\dfrac{1+\alpha}{2}$ and plug it in, the inequality follows $\Box$
