Show that the sequence $\sum\limits_{k=1}^n\cos\left(\frac kn\right)^{2n^2/k}$ converges How to prove that $$u_n=\sum_{k=1}^n\cos\left(\frac kn\right)^{2n^2/k}$$
is a Cauchy sequence?
The exercise I am reading gives as a hint that we should use the inequality: $$0\leq\cos\left(\frac kn\right)^{2n^2/k}\leq e^{-k},$$
for all $k\leq n$.
I tried to estimate $|u_m-u_n|$, but I don't know how to deal with the $n$ inside the sum.
 A: First of all,
$\cos\left(\frac{k}{n}\right)^{\frac{2n^2}{k}}\leq e^{-k}
$
is equivalent to
$\cos\left(\frac{k}{n}\right)
\leq e^{-k^2/2n^2}
$
which follows from
$e^{-x}
\ge 1-x+x^2/2$
for $\frac12 \ge x \ge 0$
(so
$e^{-k^2/2n^2}
\ge 1-k^2/(4n^2)+k^4/(8n^4)
$)
and
$\cos(x)
\le 1-x^2/2+x^4/24$
(so
$\cos(k/n)
\le 1-k^2/(2n^2)+k^4/(24n^4)
$).
Note that we have to go
to the $x^4$ term.
Then
$u_n
=\sum_{k=1}^n\cos\left(\frac{k}{n}\right)^{\frac{2n^2}{k}}
\le\sum_{k=1}^n e^{-k}
\lt \frac{1}{e-1}
$
so
$u_n$ is a bounded sequence.
However,
we have not yet shown
that
$u_n$ is increasing.
Instead
I will show that
$u_n
\to \frac{1}{e-1}
$
with too much computation.
We have
$\cos\left(\frac{k}{n}\right)^{\frac{2n^2}{k}}
\ge \left(1-k^2/(2n^2)\right)^{\frac{2n^2}{k}}
= \left(\left(1-k^2/(2n^2)\right)^{\frac{2n^2}{k^2}}\right)^k
$.
For $0 < x \le \frac12$,
$\begin{array}\\
-\ln(1-x)
&=\sum_{k=1}^{\infty} \dfrac{x^k}{k}\\
&=x+\sum_{k=2}^{\infty} \dfrac{x^k}{k}\\
&\lt x+\sum_{k=2}^{\infty} \dfrac{x^k}{2}\\
&=x+\dfrac{x^2}{2(1-x)}\\
&\le x+x^2/4\\
\end{array}
$
or
$\ln(1-x)
\ge -x-x^2/4
$.
Therefore,
for $0 < x \le \frac12$,
$(1-x)^{1/x}
=\exp((1/x)\ln(1-x))
\ge\exp(-(1/x)(x+x^2/4))
=\exp(-1-x/4)
$
so
$\left(1-k^2/(2n^2)\right)^{\frac{2n^2}{k^2}}
\ge \exp(-1-k^2/(8n^2))
$
so
$\left(\left(1-k^2/(2n^2)\right)^{\frac{2n^2}{k^2}}\right)^k
\ge \exp(-k-k^3/(8n^2))
$
so
$u_n
=\sum_{k=1}^n\cos\left(\frac{k}{n}\right)^{\frac{2n^2}{k}}
\ge\sum_{k=1}^n \exp(-k)\exp(-k^3/(8n^2))
$.
I will now split the sum
into two parts:
$[1, n^c]$
and
$(n^c, n]$
where
$0 < c < 1$.
If
$k > n^{c}$,
all the terms are positive,
so that sum is positive.
For the rest of $u_n$,
since
$e^{-x}
\gt 1-x
$,
$\begin{array}\\
\sum_{k=1}^{n^{c}} \exp(-k)\exp(-k^3/(8n^2))
&\gt \sum_{k=1}^{n^{c}} \exp(-k)(1-k^3/(8n^2))\\
&= \sum_{k=1}^{n^{c}} \exp(-k)-\sum_{k=1}^{n^{c}} \exp(-k)k^3/(8n^2)\\
&= \sum_{k=1}^{n^{c}} \exp(-k)-\sum_{k=1}^{n^{c}} \exp(-k)n^{3c-2}/8\\
&\gt \sum_{k=1}^{n^{c}} \exp(-k)-\frac{n^{3c-2}}{8}\sum_{k=1}^{n^{c}} \exp(-k)\\
\end{array}
$
and the first term
approaches
$\sum_{k=1}^{\infty} \exp(-k)
=\frac{1/e}{1-1/e}
=\frac{1}{e-1}
$
and the second term
is less than
$\frac{n^{3c-2}}{8(e-1)}
$
which goes to zero
if
$c < \frac23$.
Therefore
the sum approaches
$\frac{1}{e-1}$.
Whew.
That was harder than I 
thought it would be.
Hope that it's correct.
A: Forget Cauchy sequences. To show the convergence of the sequence $(u_n)$, consider, for every integer positive $(k,n)$, $$v_{k,n}=\cos\left(\frac{k}{n}\right)^{2n^2/k}\mathbf 1_{k\leqslant n}$$ then $$u_n=\sum_{k=1}^\infty v_{k,n}$$ Assume momentarily that, for every $x$ in $[0,1]$, $$\cos x\leqslant e^{-x^2/2}\tag{$\ast$}$$ Then, for every positive $(k,n)$, $$|v_{k,n}|=v_{k,n}\leqslant\left(e^{-k^2/(2n^2)}\right)^{2n^2/k}=e^{-k}$$ and, for every fixed positive $k$, $$\lim_{n\to\infty}v_{k,n}=e^{-k}$$ Thus, by Lebesgue dominated convergence theorem for series, $$\lim_{n\to\infty} u_n=\sum_{k=1}^\infty\lim_{n\to\infty} v_{k,n}=\sum_{k=1}^\infty e^{-k}=\frac1{e-1}$$
To complete the proof, one needs to prove $(\ast)$ but this is direct, considering the derivative of the function $$f(x)=e^{x^2/2}\cos x$$ on $[0,1]$ and using the estimate $\tan x\geqslant x$ on this interval.
A: Starting from the high school limit $\lim_{x\to 0}\log(\cos(x))/x^2=-1/2$, we have near $0$ that $\cos(x)\approx e^{-x^2/2}$. Intuitively (having in mind the Taylor series and focusing on the interval $x$ in $[0,1]$), it's clear now you can bound the sum using the exponential function as you state in your post.  
More explanations: The high school limit gives us the starting idea, and the Taylor series of $e^{-x^2/2}-\cos(x)$, with $x$ in $[0,1]$ assures the desired inequality.  
A: By making use of the inequality $0\leq cos(\frac{k}{n})^{2n^2/k}\leq e^{-k}$ for all $k\leq n$ we can state that
$$\sum_{k=1}^n cos(\frac{k}{n})^{2n^2/k}\leq\sum_{k=1}^ne^{-k}$$
If an $f$ function verifies that $f'(x)<0$ and $f''(x)>0$ for all $x\in[u,v]$ then we can apply the following inequality
$$\sum_{k=u}^vf(k)\leq\int_u^{v+1}f(x)dx+f(u)-f(v+1)$$
As the function $e^{-x}$ meets the requirements of the statement for all $x\in\mathbb{R}$ we can prove
$$\sum_{k=1}^ne^{-k}\leq\int_1^{n+1}e^{-x}dx+e^{-1}-e^{-n-1}$$
If $n\in\mathbb{N}$ it is convergent, so we just have to know what happens when $n\rightarrow\infty$. Hence,
$$\lim_{n\rightarrow\infty}\sum_{k=1}^ne^{-k}\leq\lim_{n\rightarrow\infty} \int_1^{n+1}e^{-x}dx+e^{-1}-e^{-n-1}$$
$$\lim_{n\rightarrow\infty}\sum_{k=1}^ne^{-k}\leq -e^{-x}|_1^\infty+e^{-1}=\frac{2}{e}$$
A: Hint. You have at least two options:


*

*use formulas for geometric series,

*use a comparison with the integral $\int_{-\infty}^0e^{t}dt=1$

