Bounding sum by (improper) integral I am trying to verify the following inequality that I came across while reviewing some analysis exercises online:
$$
\sum_{n=1}^{k} \left(1-\frac{n}{k}\right)n^{-1/7}\leq \int_{0}^{k}\left(1-\frac{x}{k}\right)x^{-1/7}\,dx, \hspace{3mm} k>1
$$
$\textbf{My question:}$ Why does the above inequality hold? Isn't the integral on the right an improper integral?
My idea was to justify the inequality by replacing $0$ with a small positive number and then using concavity of the function $f(x)=(1-x/k)x^{-1/7}$ but since the function is concave up I am having a hard time justifying it.
 A: Test by starting with $k=1$, then you have
$$
0\leq\int_0^1(x^{-1/7}-x^{6/7})dx=[7\frac{x^{6/7}}{6}-7\frac{x^{13/7}}{13}]_0^1=\frac{49}{78}\tag{1}
$$
which is true. Next examine the difference of the sums
$$
\Delta S_k=\sum_{n=1}^{k+1}(1-\frac{n}{k+1})n^{-1/7}-\sum_{n=1}^{k}(1-\frac{n}{k})n^{-1/7}=\\
=0+\sum_{n=1}^{k}(1-\frac{n}{k+1})n^{-1/7}-\sum_{n=1}^{k}(1-\frac{n}{k})n^{-1/7}=\\=\sum_{n=1}^{k}(1-\frac{n}{k+1}-1+\frac{n}{k})n^{-1/7}=\frac{1}{k(k-1)}\sum_{n=1}^{k}n^{6/7}
$$
with the difference of the integrals
$$
\Delta I_k=\int_0^{k+1}(1-\frac{x}{k+1})x^{-1/7}dx-\int_0^k(1-\frac{x}{k})x^{-1/7}dx=\\=\int_k^{k+1}(1-\frac{x}{k+1})x^{-1/7}dx+\frac{1}{k(k+1)}\int_0^kx^{6/7}dx>\\>\frac{1}{k(k+1)}\int_0^kx^{6/7}dx>\frac{1}{k(k-1)}\sum_{n=1}^{k}n^{6/7}=\Delta S_k
$$
because $x^{6/7}$ is at least as big as $n^{6/7}$ for $n=1,\ldots,k$ because $n$ is the integer part of $x$ for a growing function. Since $\Delta I_k>\Delta S_k$ for all $k\geq 1$ and (1) holds, the inequality in your exersice proven.
A: Setting the scene: define $f_k(x) = (1-\frac xk)x^{-1/7}$, so that the inequality in question is $\sum_{n=1}^k f_k(n) \le \int_0^k f_k(x)\,dx$. This inequality is of course equivalent (for $k\ge1$) to
$$
f_k(1) + \sum_{n=2}^k f_k(n) \le \int_0^1 f_k(x)\,dx + \int_1^k f_k(x)\,dx.
$$
Since the OP indicated that they're comfortable with the inequality $\sum_{n=2}^k f_k(n) \le \int_1^k f_k(x)\,dx$ (since $f_k$ is decreasing), let's focus on the inequality
$$
f_k(1) \le \int_0^1 f_k(x)\,dx
$$
that contains the improper integral.
The main point: I claim that the fact that the integral is improper is really a red herring. Indeed, this last inequality is equivalent to
$$
f_k(1) - f_k(1) \le \int_0^1 f_k(x)\,dx - f_k(1),
$$
or simply
$$
0 \le \int_0^1 \big( f_k(x) - f_k(1) \big) \,dx.
$$
And, improper or not, this inequality is obvious because the integrand is nonnegative (again since $f_k$ is decreasing).
A: In other words, using generalized harmonic numbers, you want to prove that
$$H_k^{\left(\frac{1}{7}\right)}-\frac{H_k^{\left(-\frac{6}{7}\right)}}{k} < \frac{49 }{78}k^{6/7}$$ For large values of $k$, using asymptotics,
$$H_k^{\left(\frac{1}{7}\right)}-\frac{H_k^{\left(-\frac{6}{7}\right)}}{k}=\frac{49 k^{6/7}}{78}+\zeta \left(\frac{1}{7}\right)-\frac{\zeta
   \left(-\frac{6}{7}\right)}{k}-\frac{1}{12}
   \left(\frac{1}{k}\right)^{8/7}+O\left(\frac{1}{k^{15/7}}\right)$$ where $\zeta \left(\frac{1}{7}\right)\sim -0.655154$ and $\zeta\left(-\frac{6}{7}\right)\sim -0.109718$.
The inequality holds as soon as $k>1$.
