Finding and proofing a closed formula for $\sum_{n=1}^k\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}$

Question

I want to find and proof a closed formula for the following sum $$\sum_{n=1}^k\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\dots +\sqrt{1+\frac{1}{k^2}+\frac{1}{(k+1)^2}}$$

I have found a closed formula but I have problems with proofing it and proofing my steps in between.

1. First I simplified the sum. To do that I calculated the radicands.
2. After that I calculated the sum for $k=1$ to $k=4$ to see a pattern.
3. Lastly I concluded from the pattern the closed formula:

$$s_k=\frac{(k+1)^2-1}{(k+1)}=(k+1)-\frac{1}{(k+1)}$$ With $s_k$ I betitle the sum up to $k$. This closed formula holds with the solutions. But now I want to proof it and my steps in between.

Simplifying the sum

I first calculated the following three radicands:

\begin{alignat*}{3} &1+\frac{1}{1^2}+\frac{1}{2^2}&&=\frac{9}{4}&&&=\frac{3^2}{2^2} \newline\newline &1+\frac{1}{2^2}+\frac{1}{3^2}&&=\frac{49}{36}&&&=\frac{7^2}{6^2} \newline\newline &1+\frac{1}{3^2}+\frac{1}{4^2}&&=\frac{169}{144}&&&=\frac{13^2}{12^2} \end{alignat*}

After calculating the radicands I saw the following pattern so that I could rewrite the sum like that: $$\sum_{n=1}^k\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=\sum_{n=1}^k\sqrt{\frac{(n\cdot(n+1)+1)^2}{(n\cdot(n+1))^2}}=\sum_{n=1}^k\frac{n^2+n+1}{n^2+n}$$

I have problems with proofing that this holds, I can't use induction to show this.

Searching for a pattern

$$ \begin{array}{c|c|c|c|c} \style{font-family:inherit}{k} &\style{font-family:inherit}{1} & \style{font-family:inherit}{2} & \style{font-family:inherit}{3} & \style{font-family:inherit}{4}\newline \hline s_k & \frac{3}{2} & \frac{8}{3} & \frac{15}{4} & \frac{24}{5} \end{array} $$

Here I calculated the sum for different $k$'s to see a pattern. I saw the correct closed formula:$$s_k=\frac{(k+1)^2-1}{(k+1)}=(k+1)-\frac{1}{(k+1)}$$

But here I am also inable to proof that that holds, as before I can't use induction.

Proofing the simplification of the sum

Because I can't use induction I tried to show that with simple transformations. I thought that I should start with showing that $1+\frac{1}{n^2}+\frac{1}{(n+1)^2}=\frac{(n^2+n+1)^2}{(n^2+n)^2}$ holds, because the removal of the square root seems trivial.

\begin{align*} &1+\frac{1}{n^2}+\frac{1}{(n+1)^2} &= \newline\newline &\frac{n^2+1}{n^2}+\frac{1}{(n+1)^2} &= \newline\newline &\frac{(n^2+1)\cdot(n+1)^2}{n^2\cdot(n+1)^2}+\frac{n^2}{n^2\cdot(n+1)^2} &= \newline\newline &\frac{(n^2+1)\cdot(n+1)^2+n^2}{n^2\cdot(n+1)^2} &= \newline\newline &\frac{n^4+2n^3+3n^2+2n+1}{n^4+2n^3+n^2} &\overset{?}{=}\newline\newline &\frac{(n^2+n+1)^2}{(n^2+n)^2} \end{align*}

I have problems in seeing the last step which is marked with the questionmark. I know that denominator is given due to the binomial theorem. And I think that also the numerator is given about the binomial theorem. But I have problems in seeing how I could transform that. You could help me with giving more smaller steps in in between the equation with the question mark. I would be able to see that this holds the other way around because that is just factorization but I have problems to see it in this way.

Proofing the closed formula

Now I want to proof the following:

$$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\dots +\sqrt{1+\frac{1}{k^2}+\frac{1}{(k+1)^2}}=(k+1)-\frac{1}{(k+1)}$$

I know from a hint that I could use the fact that this is a telescoping sum. So I rewrote the sum into the following (I can do this because of the simplification):

$$\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+\dots+\frac{k^2+k+1}{k^2+k}$$

From here I tried different expansions, but didn't came up with one where the terms cancel each other.

I tried to expand it like this:

$$(\frac{4}{2}-\frac{1}{2})+(\frac{8}{6}-\frac{1}{6})+(\frac{14}{12}-\frac{1}{12})+\dots+(\frac{k^2+k+2}{k^2+k}-\frac{1}{k^2+k})$$

and like this:

$$(2-\frac{1}{2})+(2-\frac{5}{6})+(2-\frac{11}{12})+\dots+(2-\frac{k^2+k-1}{k^2+k})$$

But in both of these tries nothing cancels.

In conclusion I would appreciate smaller steps which help at my proof of the simplification of the sum, and I would appreciate help in finding the expansion such that the terms cancel.

Answer 1

Note that :

$$\frac{n^2+n+1}{n^2+n}=\frac{n(n+2)}{n+1}-\frac{(n-1)(n+1)}{n}$$

Therfore: $$\sum_{n=1}^k\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}} =\sum_{n=1}^k\frac{n^2+n+1}{n^2+n}=\sum_{n=1}^k \left({\frac{n(n+2)}{n+1}-\frac{(n-1)(n+1)}{n}}\right)=\frac{k(k+2)}{k+1}=k+1-\frac{1}{k+1}$$

Answer 2

$$S :=\sum_{n=1}^k\sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}$$ $$= \sum_{n=1}^k\sqrt{\frac{n^2(n+1)^2+(n+1)^2+n^2}{n^2(n+1)^2}}$$ $$= \sum_{n=1}^k{\frac{\sqrt{n^4+2n^3+n^2+n^2+2n+1+n^2}}{n(n+1)}}$$ $$= \sum_{n=1}^k{\frac{\sqrt{n^4+2n^3+3n^2+2n+1}}{n(n+1)}}$$

The radicand has no rational roots, but maybe we can factor it into two quadratics.

$$n^4+2n^3+3n^2+2n+1 = (n^2 + an + b)(n^2 + cn + d)$$ $$n^4+2n^3+3n^2+2n+1 = n^4 + cn^3 + dn^2 + an^3 + acn^2 + adn + bn^2 + bcn + bd$$ $$n^4+2n^3+3n^2+2n+1 = n^4 + (a + c)n^3 + (ac + b + d)n^2 + (ad + bc)n + bd$$

Matching up the coefficients gives the system of equations:

$$a + c = 2 \tag{1}$$ $$ac + b + d = 3\tag{2}$$ $$ad + bc = 2 \tag{3}$$ $$bd = 1 \tag{4}$$

The only integer solutions to (4) are $b = d = 1$ and $b = d = -1$.

If $b = d = 1$, then (1) and (3) both become $a + c = 2$, and (2) becomes $ac = 1$.

If $b = d = -1$, then (3) becomes $a + c = -2$, which contradicts equation (1), $a + c = 2$.

So we must have $b = d = 1$. From (1), $c = 2 - a$, and plugging into (2) gives the quadratic equation $a^2 - 2a + 1 = 0$, or $(a - 1)^2 = 0$. So $a = 1$, so $c = 1$, and thus:

$$n^4+2n^3+3n^2+2n+1 = (n^2 + n + 1)(n^2 + n + 1)$$ $$\sqrt{n^4+2n^3+3n^2+2n+1} = n^2 + n + 1$$

This takes care of the $\overset{?}{=}$ step in your question. Substituting this expression into $S$, we get:

$$S = \sum_{n=1}^k{\frac{n^2 + n + 1}{n(n+1)}}$$ $$= \sum_{n=1}^k \left(\frac{n^2 + n}{n(n+1)} + \frac{1}{n(n+1)}\right)$$ $$= \sum_{n=1}^k \left(1 + \frac{1}{n(n+1)}\right)$$ $$= k + \sum_{n=1}^k \frac{1}{n(n+1)}$$

Let's do a partial fraction decomposition on the summand.

$$\frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1}$$ $$1 = A(n+1) + Bn$$ $$1 = (A + B)n + A$$

Matching up the coefficients gives us the system of equations $A+B=0$ and $A=1$, and thus $B=-1$. So, $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.

$$S = k + \sum_{n=1}^k \left( \frac{1}{n} - \frac{1}{n+1} \right)$$ $$= k + \sum_{n=1}^k \frac{1}{n} - \sum_{n=1}^k \frac{1}{n+1}$$

Let $m = n$ in the first summation, and $m = n + 1$ in the second summation.

$$S= k + \sum_{m=1}^k \frac{1}{m} - \sum_{m=2}^{k+1} \frac{1}{m}$$

Note that the terms from $m = 2$ to $k$ cancel out, and so we are left with:

$$S= k + \sum_{m=1} \frac{1}{m} - \sum_{m=k+1} \frac{1}{m}$$ $$= k + 1 - \frac{1}{k+1}$$

Answer 3

The first equality

$$n^4+2n^3+3n^2+2n+1 = (n^2+n+1)^2$$

can be proved to be true by just expanding the right side. It's also obvious if you are familiar with convolutions.

Answer 4

$$1+\dfrac1{(n+a)^2}+\dfrac1{(n+b)^2}=\cdots=\dfrac{(n^2+(a+b)n+ab)^2+2n^2+2n(a+b)+a^2+b^2}{(n+a)^2(n+b)^2}$$

Comparing the numerator with

$$(n^2+cn+d)^2=n^4+2cn^3+n^2(c^2+2d)+2n(cd)+d^2$$

Equating the constants and the coefficients of $n^3,n^2,n$

$c=a+b$

and $c^2+2d=(a+b)^2+2=c^2+2\implies d=1$

and $(a+b)(1+ab)=cd\iff c(1+ab)=c\iff ab=0$

$\implies$ either $a=0\implies c=b$

and $a^2+b^2+a^2b^2=d^2=1\implies b=\pm 1$

$\implies (n^2\pm n+1)^2=n^4\pm2n^3+3n^2\pm2n+1$

Similarly if $b=0$

Now if $g(n)=\dfrac{n^2+n+1}{n(n+1)}=1+\dfrac{n+1-n}{n(n+1)}=1+f(n)-f(n+1)$ where $f(m)=\dfrac1m$

$$\sum_{n=1}^kg(n)=\sum_{n=1}^k1+\sum_{n=1}^k\underbrace{(f(n)-f(n+1))}_{\text{ Telescoping series}}=k + f(1)-f(k+1)$$