MHB Evaluating Summation: Find $\sum_{i=1}^{\infty} (-1)^{i+1}f(i)$

[Saitama]

Problem:
Consider a function $f(n)$ defined as:
$$f(n)=\sum_{r=1}^n (-1)^{r+1} \binom{n}{r} \left(\sum_{k=1}^r \frac{1}{k}\right)$$
Find the value of
$$\sum_{i=1}^{\infty} (-1)^{i+1}f(i)$$

Attempt:
I write $\sum_{k=1}^r (1/k)=H_r$.

The sum I have to evaluate is
$$f(1)-f(2)+f(3)-f(4)+\cdots$$

I tried writing down a few terms and tried to see the difference of consecutive terms...
$$f(1)=H_1$$
$$f(2)=2H_1-H_2$$
$$f(3)=3H_1-3H_2+H_3$$
$$f(4)=4H_1-6H_2+4H_3+H_4$$
...but I don't see if this helps.

Although I have posted this in the Pre-Algebra and Algebra forum, please feel free to use any Calculus approaches as I am not sure if the problem involves the use of Calculus.

Any help is appreciated. Thanks!

 

[chisigma]

Problem:
Consider a function $f(n)$ defined as:
$$f(n)=\sum_{r=1}^n (-1)^{r+1} \binom{n}{r} \left(\sum_{k=1}^r \frac{1}{k}\right)$$
Find the value of
$$\sum_{i=1}^{\infty} (-1)^{i+1}f(i)$$

Attempt:
I write $\sum_{k=1}^r (1/k)=H_r$.

The sum I have to evaluate is
$$f(1)-f(2)+f(3)-f(4)+\cdots$$

I tried writing down a few terms and tried to see the difference of consecutive terms...
$$f(1)=H_1$$
$$f(2)=2H_1-H_2$$
$$f(3)=3H_1-3H_2+H_3$$
$$f(4)=4H_1-6H_2+4H_3+H_4$$
...but I don't see if this helps.

Although I have posted this in the Pre-Algebra and Algebra forum, please feel free to use any Calculus approaches as I am not sure if the problem involves the use of Calculus.

Any help is appreciated. Thanks!

Your approach is very good because is... $f(1) = H_{1} = 1$

$f(2) = 2\ H_{1} - H_{2} = 2 - 1 - \frac{1}{2} = \frac{1}{2}$

$f(3) = 3\ H_{1} - 3\ H_{2} + H_{3} = 3 - 3 - \frac{3}{2} + 1 + \frac{1}{2} + \frac{1}{3} = \frac{1}{3}$ ... and proceeding in this way You arrive at the very suggestive relation...

$\displaystyle f(n) = \sum_{k=1}^{n} (-1)^{k+1} \binom{n}{k}\ H_{k} = \frac{1}{n}\ (1)$... so that is...$\displaystyle \sum_{i=1}^{\infty} (-1)^{i + 1} f(i) = \ln 2\ (2)$

Kind regards

$\chi$ $\sigma$

 

[polygamma]

There is a proof that $ \displaystyle H_{n} = \sum_{k=1}^{n}(-1)^{k-1} \binom{n}{k} \frac{1}{k} $ on Wikipedia page for the harmonic numbers.

Harmonic number - Wikipedia, the free encyclopediaThen by applying the inverse binomial transform,

$$ \frac{1}{n} = \sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} H_{k} $$Binomial Transform -- from Wolfram MathWorld

 

[Prometheus1]

Switch the order of the inner sums and use $\displaystyle \sum\limits_{k \le r \le n} (-1)^r\binom{n}{r} = \frac{k (-1)^k}{n} \binom{n}{k}.$

$$\displaystyle \begin{aligned} \sum_{n \ge 1}\sum_{1 \le r \le n} \sum_{1 \le k \le r} \frac{(-1)^{n+r}}{k} \binom{n}{r} & = \sum_{n \ge 1}\sum_{1 \le k \le n} \sum_{k \le r \le n} \frac{(-1)^{n+r}}{k} \binom{n}{r} \\& = \sum_{n \ge 1}\sum_{1 \le k \le n}\frac{(-1)^{k+n}}{n} \binom{n}{k} \\& = \sum_{n \ge 1} \frac{(-1)^{n+1}}{n} \\& = \log(2).\end{aligned} $$​

 

[polygamma]

@Prometheus

Could you give a proof of that identity?

 

[Saitama]

Your approach is very good because is... $f(1) = H_{1} = 1$

$f(2) = 2\ H_{1} - H_{2} = 2 - 1 - \frac{1}{2} = \frac{1}{2}$

$f(3) = 3\ H_{1} - 3\ H_{2} + H_{3} = 3 - 3 - \frac{3}{2} + 1 + \frac{1}{2} + \frac{1}{3} = \frac{1}{3}$ ... and proceeding in this way You arrive at the very suggestive relation...

$\displaystyle f(n) = \sum_{k=1}^{n} (-1)^{k+1} \binom{n}{k}\ H_{k} = \frac{1}{n}\ (1)$... so that is...$\displaystyle \sum_{i=1}^{\infty} (-1)^{i + 1} f(i) = \ln 2\ (2)$

Kind regards

$\chi$ $\sigma$

Awesome!

I never thought I was so close to the answer, thanks a lot chisigma! :)

There is a proof that $ \displaystyle H_{n} = \sum_{k=1}^{n}(-1)^{k-1} \binom{n}{k} \frac{1}{k} $ on Wikipedia page for the harmonic numbers.

Harmonic number - Wikipedia, the free encyclopediaThen by applying the inverse binomial transform,

$$ \frac{1}{n} = \sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} H_{k} $$Binomial Transform -- from Wolfram MathWorld

Hi Random Variable, thank you very much for your input. :)

I have never heard of inverse binomial transform before but it looks like a useful technique. I seem to have trouble figuring out how did you use it here.

We have
$$H_{n} = \sum_{k=1}^{n}(-1)^{k-1} \binom{n}{k} \frac{1}{k} $$
From the Wolfram page you quote, $b_n$ should be of the form $\displaystyle \sum_{k=0}^n (-1)^{n-k} a_k$ but I have $(-1)^{k-1}$ instead of $(-1)^{n-k}$.

Sorry if this is silly but I have never seen that representation of $H_n$ and the binomial transform.

Thanks!

 

[polygamma]

@ PranavFirst let $a_{0}=b_{0} = 0 $.

Then multiply both sides of $ \displaystyle H_{n} = \sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} \frac{1}{k} $ by $(-1)^{n+1}$.

Then $ \displaystyle (-1)^{n+1} H_{n} = (-1)^{n-1} H_{n} = \sum_{k=1}^{n} (-1)^{n+k} \binom{n}{k} \frac{1}{k} = \sum_{k=1}^{n} (-1)^{n-k} \binom{n}{k} \frac{1}{k}$.

Now take the inverse binomial transform.

 

[Saitama]

@ PranavFirst let $a_{0}=b_{0} = 0 $.

Then multiply both sides of $ \displaystyle H_{n} = \sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} \frac{1}{k} $ by $(-1)^{n+1}$.

Then $ \displaystyle (-1)^{n+1} H_{n} = (-1)^{n-1} H_{n} = \sum_{k=1}^{n} (-1)^{n+k} \binom{n}{k} \frac{1}{k} = \sum_{k=1}^{n} (-1)^{n-k} \binom{n}{k} \frac{1}{k}$.

Now take the inverse binomial transform.

Thanks a lot Random Variable! :)