Conjectured formula to take a derivative out of a summation

Question

Does anyone know of a proof or disproof of this conjectured formula?

$\sum_{k=1}^{n}\frac{\mathrm{d} }{\mathrm{d} k}f(k)=C+\frac{\mathrm{d} }{\mathrm{d} n}\sum_{k=1}^{n}f(k)$

I was working on evaluating this infinite sum from a question that I found:

$\sum_{k=1}^{\infty}2^{-k}\tan(2^{-k})$

While working on evaluating that infinite sum, I have encountered the issue of taking derivatives out of a summation, when expressing $\tan(2^{-k})$ as $\frac{-\frac{\mathrm{d} }{\mathrm{d} k}\ln\cos(2^{-k})}{\frac{\mathrm{d} }{\mathrm{d} k}2^{-k}}$ (using the chain rule with $\frac{\mathrm{d} }{\mathrm{d} x}\ln\cos x = -\tan x$). Taking the derivative out (after cancelling out the $2^{-k}$ in the summation with the one from the derivative) allows you to turn the summation into the repeated product of $\cos(2^{-k})$ inside of the $\ln$.

You can use a formula obtained from the repeated application of the double-angle formula, taking x to be 1, and rearrange it:

$\sin x = 2^{n}\sin\frac{x}{2^{n}}\prod_{k=1}^{n}\cos\frac{x}{2^{k}}$

$\prod_{k=1}^{n}\cos(2^{-k}) = (\frac{2^{-n}}{sin(2^{-n})})\sin 1$

Which finally allows you to evaluate the infinite sum by applying this formula on the repeated product

The issue is that, to take the derivative out of the summation, I had to rely on this conjectured formula that I made, that I wasn't able to prove, but also wasn't able to find any counter-examples to, including that the solution I arrived at, $\frac{1}{2}(\tan(\frac{1}{2})-\cot(\frac{1}{2})+2)$, which seems to be accurate from the estimations of this infinite sum to 6 digits.

Answer 1

Given $f(x)\in C^{1}(\mathbb{Z}_{+})$ (by this I mean its value and derivative at all positve integer exists). Let $g(x)$ is a smooth function satisfying $h(x)=0\ \forall x\in\mathbb{Z}_{+}$ (*) where $h(x)=g(x)-g(x-1)-f(x)$, this implies $\sum_{k=1}^{n}f(k)=g(n)-g(0)$.
To construct such $g(x)$ so that the reformulated conjecture is true, we must have $h(x)\in C^{1}(\mathbb{Z}_{+})/H$ where
$$\forall y(x)\in H:(y(x)=0,\ \forall x\in\mathbb{Z}_{+})\wedge(\exists x\in\mathbb{Z}_{+}:y'(x)\not =0)$$ This quotient space ensures that: $$h(x)=0\ \forall x\in\mathbb{Z}_{+}\implies h'(x)=0\ \forall x\in\mathbb{Z}_{+}$$ Consequently, all "pseudo-zeros" like $\sin(\pi x)$ never affect our results.
Now that we have $h'(x)=g'(x)-g'(x-1)-f'(x)=0\ \forall x\in\mathbb{Z}_{+}$. OP's informal statement: $$\sum_{k=1}^{n}\frac{d}{dk}f(k)=C+\frac{d}{dn}\sum_{k=1}^{n}f(k)\tag{1}$$ can be restated as $$\sum_{k=1}^{n}f'(k)=\sum_{k=1}^{n}(g'(k)-g'(k-1))=-g'(0)+g'(n)\tag{2}$$
where $f'(k)$ replaces $\frac{d}{dk}f(k), -g'(0)$ replaces $C$, $g'(n)$ replaces $\frac{d}{dn}\sum_{k=1}^{n}f(k)$.
OP's conjecture $(1)$ is blatantly false but its reformulated version $(2)$ is true.

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Just for fun, one may wish to define "derivative on integers" by temporarily transforming domain:
$$\mathbb{Z}_{+}\xrightarrow{\textrm{interpolate}}\mathbb{R}\xrightarrow{\textrm{collapse}}\mathbb{Z}_{+}$$ I will leave this idea open.

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Note: The function space $C^{1}(\mathbb{Z}_{+})/H$ will never be null and we will always be able to find $g(x)$. Just for example, take modified Whittaker–Shannon interpolation:
$$g(x)=\sum_{n=1}^{\infty}\left(\operatorname{sinc}(x-n)^2\sum_{k=1}^{n}f(k)+\operatorname{sinc}(x-n)\int_{0}^{x-n}\operatorname{sinc}(t)dt\sum_{k=1}^{n}f'(k)\right)$$
Note that all those $\operatorname{sinc}(x)$ are normalized and this only hold for a certain class of functions $f(x)$.
When $f(x)$ is a polynomial or an exponential, we can easily choose $g(x)$, just remember $g(x)$ is never necessarily unique, it depends on how we want to interpolate.

Answer 2

Take $f(x) = \sin(2 \pi x)$

Then $$\sum_{k=1}^{n}\frac{d}{d k}f(k)=\sum_{k=1}^{n}2\pi\cos(2\pi k)=2\pi n$$ while $$\frac{d}{dn}\sum_{k=1}^{n}f(k)=\frac{d}{dn}\sum_{k=1}^{n}0=0$$

This two functions does not differ by a constant.

Answer 3

As Thomas Andrews suggested in a comment, we can verify this claim concretely when treating everything as a polynomial.

Suppose $f(x)=\sum_{j=0}^{m}a_{j}x^{j}$. Then $\sum_{k=1}^{n}f'(k)=\sum_{k=1}^{n}\sum_{j=0}^{m}ja_{j}k^{j-1}=\sum_{k=1}^{n}\sum_{j=1}^{m}ja_{j}k^{j-1}=\sum_{j=1}^{m}ja_{j}\sum_{k=1}^{n}k^{j-1}$. We can apply Faulhaber's formula (involving the Bernoulli numbers $B_r$) to write this in terms of powers of $n$ instead of $k$, and obtain $$\sum_{j=1}^{m}ja_{j}\left(\dfrac{1}{j-1+1}\right)\sum_{r=0}^{j-1}{j-1+1 \choose r}B_{r}n^{j-1-r+1}=\boxed{\sum_{j=1}^{m}a_{j}\sum_{r=0}^{j-1}{j \choose r}B_{r}n^{j-r}}\text{.}$$

Similarly, we can calculate $\sum_{k=1}^{n}f(k)=\sum_{k=1}^{n}\sum_{j=0}^{m}a_{j}k^{j}=\sum_{k=1}^{n}\sum_{j=1}^{m}a_{j}k^{j}=\sum_{j=1}^{m}a_{j}\sum_{k=1}^{n}k^{j}$ and apply Faulhaber's formula to obtain $\sum_{j=1}^{m}a_{j}\left(\dfrac{1}{j+1}\right)\sum_{r=0}^{j}{j+1 \choose r}B_{r}n^{j-r+1}$.

And then since this is now written as a polynomial in $n$, we can take $\dfrac{\mathrm{d}}{\mathrm{d}n}$ of it in a straightforward way (ignoring any non-polynomials that happen to equal it for nonnegative integers), to obtain: \begin{align*}\phantom{=}&\sum_{j=1}^{m}a_{j}\left(\dfrac{1}{j+1}\right)\sum_{r=0}^{j}{j+1 \choose r}\left(j-r+1\right)B_{r}n^{j-r}\\=&\sum_{j=1}^{m}a_{j}\left(\dfrac{1}{j+1}\right)\sum_{r=0}^{j}\left(j+1\right){j \choose r}B_{r}n^{j-r}\\=&\sum_{j=1}^{m}a_{j}\sum_{r=0}^{j}{j \choose r}B_{r}n^{j-r}\\=&\sum_{j=1}^{m}a_{j}\left(B_{j}+\sum_{r=0}^{j-1}{j \choose r}B_{r}n^{j-r}\right)\\=&\sum_{j=1}^{m}a_{j}B_{j}+\boxed{\sum_{j=1}^{m}a_{j}\sum_{r=0}^{j-1}{j \choose r}B_{r}n^{j-r}}\\=&\sum_{j=1}^{m}a_{j}B_{j}+\sum_{k=1}^{n}f'(k)\text{.}\end{align*}

Therefore, in the case of polynomials, the claim is true with the constant $C=\sum_{j=1}^{m}a_{j}B_{j}$.

However, we cannot naively extend this result from polynomials to a common power series like that of $\cos(k)$, as the Bernoulli numbers grow pretty fast. There may be an argument about functions that decay very fast like those functions with $2^{-k}$ in them in the OP, but that is out of my wheelhouse and would be deserving of its own separate question.