How to prove that this sum of binomials is increasing? Let
$$
S_k = {1\over 2^{3 k}} \sum_{j=k}^{3 k} {3 k \choose j}.
$$
I checked the values $S_1,\ldots,S_{100}$ and they seem to be increasing (and converge to $1$). Is it possible to prove that the sequence $S_k$ is always increasing? 
 A: Convergence to $1$ is straightforward through the central limit theorem and the Berry-Esseen theorem. 
Proving that $S_k$ is increasing is equivalent to proving that the following is decreasing:
\begin{align*}
T_k &= \frac{1}{8^k} \sum_{j=0}^{k-1}\binom{3k}{j}
\\
    &= \frac{1}{8^k} \sum_{j=0}^{k-1}\sum_{j'=0}^{3 k}\binom{3k}{j'} {1\over 2\pi}\int_{\theta=0}^{2\pi} e^{-j i \theta} e^{j' i \theta} \,d\theta 
&& \text{(since the integral equals 1 iff $j=j'$)}
\\
    &= \frac{1}{2\pi \cdot 8^k}\int_{\theta=0}^{2\pi}
\sum_{j=0}^{k-1} e^{-j i \theta} \sum_{j'=0}^{3 k}\binom{3k}{j'} e^{j' i \theta}
 \,d\theta 
\\
    &= \frac{1}{2\pi \cdot 8^k}\int_{\theta=0}^{2\pi}
\sum_{j=0}^{k-1} e^{-j i \theta} (1 + e^{i \theta})^{3 k}
 \,d\theta 
&& \text{(by the Binomial theorem)}
\\
    &= \frac{1}{2\pi \cdot 8^k}\int_{\theta=0}^{2\pi}
{1 - e^{- i k \theta} \over 1 - e^{- i \theta}} (1 + e^{i \theta})^{3 k}
 \,d\theta 
&& \text{(sum of a finite geometric series)}
\\
 & = \frac{1}{2\pi\cdot 8^k}\int_{0}^{2\pi}\frac{(1+e^{i\theta})^{3k}\left(1-e^{-ik\theta}\right)}{1-e^{-i\theta}}\,d\theta 
\end{align*}
$T_k$ is clearly a positive sequence. 
By the Cauchy-Schwarz inequality, it is possible to prove that it is log-convex, i.e, for all $k$:
\begin{align*}
(*)
&&
\left(
\int_{0}^{2\pi}\frac{(1+e^{i\theta})^{3(k-1)}\left(1-e^{-i(k-1)\theta}\right)}{1-e^{-i\theta}}\,d\theta 
\right)
\left(
\int_{0}^{2\pi}\frac{(1+e^{i\theta})^{3(k+1)}\left(1-e^{-i(k+1)\theta}\right)}{1-e^{-i\theta}}\,d\theta 
\right)
\\
\geq &&
\left(
\int_{0}^{2\pi}\frac{(1+e^{i\theta})^{3k}\left(1-e^{-ik\theta}\right)}{1-e^{-i\theta}}\,d\theta 
\right)
\left(
\int_{0}^{2\pi}\frac{(1+e^{i\theta})^{3k}\left(1-e^{-ik\theta}\right)}{1-e^{-i\theta}}\,d\theta 
\right)
\end{align*}
which is equivalent to:
\begin{align*}
(**)
&&
T_{k-1}\cdot T_{k+1}  \geq T_{k}^2
\end{align*}
This log-convexity and the fact $T_k\to 0$ imply that $\{T_k\}_{k\geq 0}$ is decreasing. Proof: Suppose by contradiction that it is not decreasing, so for some $k$, $T_k \geq T_{k-1}$. Then, by (**):
$$
T_{k+1} \geq T_{k} {T_k\over  T_{k-1}} \geq T_k
$$
so the sequence $T_k$ is bounded from below by $T_{k-1}$ and cannot converge to 0 - a contradiction.
