Sum with binomial coefficients: $\sum_{m=1}^{k}\frac{1}{m^{2a}}\binom{k}{m}$ with constant a

Question

I don’t know how to give the upper bound of the following formula (the upper bound is related to $k$, and I hope the order of $k$ is a negative number)

$$ \sum_{m=1}^{k}\frac{1}{m^{2a}}\binom{k}{m} $$ where $a\in (1/2, 1]$ is a constant. Any help in evaluating this sum would be appreciated, thanks!

Answer 1

New Solution. I shall derive a complete asymptotic expansion for large $k$. I will assume that $a>0$ is fixed. Using $$ \frac{1}{{m^{2a} }} = \frac{1}{{\Gamma (2a)}}\int_0^{ + \infty } {{\rm e}^{ - mt} t^{2a - 1} {\rm d}t} , $$ we readily find that $$ \sum\limits_{m = 1}^k {\frac{1}{{m^{2a} }}} \binom{k}{m} = \frac{1}{{\Gamma (2a)}}\int_0^{ + \infty } {((1 + {\rm e}^{ - t} )^k - 1)t^{2a - 1} {\rm d}t} . $$ Integrating once by parts gives $$ \sum\limits_{m = 1}^k {\frac{1}{{m^{2a} }}} \binom{k}{m} =\frac{k}{{\Gamma (2a + 1)}}\int_0^{ + \infty } {(1 + {\rm e}^{ - t} )^{k - 1} {\rm e}^{ - t} t^{2a} {\rm d}t} . $$ It is readily seen that $$ \frac{k}{{\Gamma (2a + 1)}}\int_{\log 2}^{ + \infty } {(1 + {\rm e}^{ - t} )^{k - 1} {\rm e}^{ - t} t^{2a} {\rm d}t} \le k\left( {\frac{3}{2}} \right)^{k - 1} = o\!\left( {\frac{{2^k }}{{k^r }}} \right) $$ as $k\to+\infty$ with any $r>0$. Accordingly, $$ \sum\limits_{m = 1}^k {\frac{1}{{m^{2a} }}} \binom{k}{m} = \frac{k}{{\Gamma (2a + 1)}}\int_0^{\log 2} {(1 + {\rm e}^{ - t} )^{k - 1} {\rm e}^{ - t} t^{2a} {\rm d}t} + o\!\left( {\frac{{2^k }}{{k^r }}} \right) $$ as $k\to+\infty$ with any $r>0$. Performing the change of integration variables from $t$ to $s$ via $2{\rm e}^{ - s} = 1 + {\rm e}^{ - t}$ yields $$ \sum\limits_{m = 1}^k {\frac{1}{{m^{2a} }}} \binom{k}{m} =\frac{{2{}^{k + 2a}k}}{{\Gamma (2a + 1)}}\int_0^{\log (4/3)} {{\rm e}^{ - ks} s^{2a} \left( {\frac{1}{{2s}}\log \left( {\frac{1}{{2{\rm e}^{ - s} - 1}}} \right)} \right)^{2a} {\rm d}s} + o\!\left( {\frac{{2^k }}{{k^r }}} \right) . $$ Then, by Watson's lemma, $$ \sum\limits_{m = 1}^k {\frac{1}{{m^{2a} }}} \binom{k}{m} \sim \frac{{2^{k + 2a} }}{{k^{2a} }}\left( {1 + \frac{{2a^2 + a}}{k} + \frac{{4a^4 + 12a^3 + 11a^2 + 3a}}{{2k^2 }} + \ldots } \right) $$ as $k\to+\infty$.

Old Solution. I will derive an asymptotics, but without going into details. Assume that $a \geq 0$ is fixed. The main contributation to the sum will come from those $m$'s for wich $\left| {k/2 - m} \right| = o(k^{2/3} )$. In this range, we can use the normal approximation for the binomials, to deduce $$ \sum_{m=1}^{k}\frac{1}{m^{2a}}\binom{k}{m} \sim \frac{{2^{k + 1/2} }}{{\sqrt {k\pi } }}\sum\limits_{\left| {k/2 - m} \right| = o(k^{2/3} )} {\frac{1}{{m^{2a} }}\mathrm{e}^{ - (k - 2m)^2 /(2k)} } $$ as $k\to +\infty$. Next, we approximate the sum by an integral and extend the range of integration to $(1,+\infty)$. This does not change the leading order behaviour. Thus, \begin{align*} \sum_{m=1}^{k}\frac{1}{m^{2a}}\binom{k}{m} &\sim \frac{{2^{k + 1/2} }}{{\sqrt {k\pi } }}\int_1^{ + \infty } {\frac{1}{{x^{2a} }}\mathrm{e}^{ - (k - 2x)^2 /(2k)} \mathrm{d}x} \\ & = \frac{{2^{k + 1/2} }}{{\sqrt \pi }}\frac{1}{{k^{2a - 1/2} }}\int_{1/k}^{ + \infty } {\frac{1}{{t^{2a} }}\mathrm{e}^{ - k(2t - 1)^2 /2} \mathrm{d}t} \end{align*} as $k\to +\infty$. Now, $$ \int_{1/k}^{1/4} {\frac{1}{{t^{2a} }}\mathrm{e}^{ - k(2t - 1)^2 /2} \mathrm{d}t} < k^{2a} \mathrm{e}^{ - k/8} $$ and we will see that this is negligible compared to the leading order behaviour of the full integral for large $k$. Hence, $$ \sum_{m=1}^{k}\frac{1}{m^{2a}}\binom{k}{m} \sim \frac{{2^{k + 1/2} }}{{\sqrt \pi }}\frac{1}{{k^{2a - 1/2} }}\int_{1/4}^{ + \infty } {\frac{1}{{t^{2a} }}\mathrm{e}^{ - k(2t - 1)^2 /2} \mathrm{d}t} . $$ Finally, we apply Laplace's method to the right-hand side to derive $$ \sum_{m=1}^{k}\frac{1}{m^{2a}}\binom{k}{m} \sim \frac{{2^{k + 2a} }}{{k^{2a} }} $$ as $k\to +\infty$.

Answer 2

When $2a$ is an integer, there is a solution in terms of generalized hypergeometric functions (but remember that they are infinite summations). Unfortunately this is not yous case.

Hoping that somebody else could help, the only thing I can write for the time being is $$k \,\, _4F_3(1,1,1,1-k;2,2,2;-1)<\sum_{m=1}^{k}\frac{1}{m^{2a}}\binom{k}{m}<k \,\, _3F_2(1,1,1-k;2,2;-1)$$ On a logarithmic scale, both lhs and rhs are very close to linearity.

Curve-fitted for $10 \leq k \leq 1000$ we have

• for the logarithm of the rhs (corresponding to $\alpha=\frac 12$), $(R^2 >0.999999)$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & -3.82501 & 0.02492 & \{-3.87392,-3.77611\} \\ b & +0.69032 & 0.00004 & \{+0.69023,+0.69040\} \\ \end{array}$$

• for the logarithm of the lhs (corresponding to $\alpha=1$), $(R^2 >0.999996)$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & -7.63288 & 0.05058 & \{-7.73214,-7.53362\} \\ b & +0.68746 & 0.00009 & \{+0.68729,+0.68763\} \\ \end{array}$$

I suppose that you notice that both slopes are very close to $\log(2)$.

What to do for intermediate values of $a$ ???

Would a log-linear interpolation would work ? In any manner, notice that $\frac{\text{rhs}}{\text{lhs}} \sim \frac k 2$

Edit

Starting from @Gary's comments and suggestions, I repeated the fits (now for the range $10 \leq k \leq 5000$ $(\Delta k=10)$ based on the model $$\log(S_a)= \alpha + \beta \,k-\gamma \, \log(k)$$ Below are reported the results (for both cases $R^2 > 0.999999999$)

• For $a=\frac 12$

$$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ \alpha & 0.757923866 & 0.002918 & \{0.752190,0.763657\} \\ \beta & 0.693151797 & 3.2 \times 10^{-7} & \{0.693151,0.693152\} \\ \gamma & 1.009949712 & 0.000477 & \{1.009012,1.010888\} \\ \end{array}$$

• For $a=1$

$$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ \alpha & 1.603545820 & 0.010736 & \{1.582452,1.624639\} \\ \beta & 0.693162912 & 1.2 \times 10^{-6} &\{0.693161,0.693165\} \\ \gamma & 2.033490830 & 0.001756 & \{2.030040,2.036942\} \\ \end{array}$$

So, it is reasonable to think that $\beta=\log(2)$ and $\gamma=2a$. What is less clear is $\alpha$ (the ratio stays close to $2$)

To try to clarify, I looked at the value of $$T_a=\log(S_a)-\big[k \log(2)-2a \log(k)\big]$$

The statistics are $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a=\frac 12 & 0.694613459 & 0.000317 & \{0.693991,0.695236\} \\ a=1 & 1.390925202 & 0.001114 & \{1.388737,1.393113\} \\ \\ \end{array}$$ In other words $\alpha=2 a \log(2)$

All of this gives an incredible weight to @Gary's asymptotics (we have the same formula).