$\frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+3+\dots+x}=\frac{2011}{2013}$

Question

I want to see OTHER approaches than this one. Make sure they are significantly different and not a direct restatement.

$$\frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+3+\dots+x}=\frac{2011}{2013}\tag{1}$$

$$\sum_{n=1}^x n=\frac{x(x+1)}{2} \; \forall x >0\tag{2}$$

$$\begin{align*} (1)&\stackrel{(2)}{\iff} \frac{2}{2\cdot 3}+\frac{2}{3\cdot 4}+\frac{2}{4\cdot 5}+\dots+\frac{2}{x(x+1)}=\frac{2011}{2013}\\\\ &\iff 2\left (\frac12 -\frac13+\frac13-\frac14+\frac14-\dots+\frac{1}{x}-\frac{1}{x+1}\right )=\frac{2011}{2013}\\\\ &\iff 1-\frac{2}{x+1}=\frac{2011}{2013}\\\\ &\iff x=2012 \end{align*}$$

Answer 1

It surely isn't significantly different, but it can be done as follows:

The first step is the same: $$ \sum_{n=1}^{x} n=\frac{x(x+1)}{2}=\binom{x+1}{2}\implies \frac{1}{1+2}+\frac{1}{1+2+3}+\dots+\frac{1}{1+2+3+\dots+x}=\sum_{n=2}^{x}\frac{1}{\binom{n+1}{2}}=\sum_{n=1}^{x-1}\frac{1}{\binom{n+2}{2}} $$ Now, take a look at the following integral: $$ \Delta(a,b)=\int_{0}^1 \left(1-t^{a}\right)^bdt $$ With the substitution $t\to t^{\frac{1}{a}}$ we get: $$ \Delta(a,b)=\int_{0}^1 \frac{1}{a}t^{\frac{1}{a}-1}\left(1-t\right)^b dt=\frac{1}{a}\frac{\Gamma\left(\frac{1}{a}\right)\Gamma(b+1)}{\Gamma\left(\frac{1}{a}+b+1\right)}=\frac{\Gamma\left(\frac{1}{a}+1\right)\Gamma(b+1)}{\Gamma\left(\frac{1}{a}+b+1\right)} $$ For $a>0$ and $b>-1$. Now, setting $a=\frac{1}{2}$ and $b=n$ we get: $$ \Delta\left(\frac{1}{2},n\right)=\int_{0}^1 \left(1-\sqrt{t} \right)^n dt=\frac{\Gamma\left(2+1\right)\Gamma(n+1)}{\Gamma\left(2+n+1\right)}=\frac{2!\cdot n!}{(n+2)!}=\frac{1}{\binom{n+2}{2}} $$ And therefore: $$ \sum_{n=1}^{x-1}\frac{1}{\binom{n+2}{2}}=\sum_{n=1}^{x-1}\left(\int_{0}^1 \left(1-\sqrt{t} \right)^n dt\right)=\int_{0}^1 \sum_{n=1}^{x-1}\left(1-\sqrt{t} \right)^n dt=\int_{0}^1 \left(1-\sqrt{t} \right)\frac{1-\left(1-\sqrt{t} \right)^{x-1}}{1-\left(1-\sqrt{t} \right)} dt=\int_{0}^1 \frac{\left(1-\sqrt{t} \right)-\left(1-\sqrt{t} \right)^{x}}{\sqrt{t}} dt $$ Now, by making the substitution $t\to(1-t)^2$ we obtain: $$ \sum_{n=1}^{x-1}\frac{1}{\binom{n+2}{2}}=\int_{1}^0 -2(1-t)\frac{\left(1-(1-t) \right)-\left(1-(1-t) \right)^{x}}{(1-t)} dt=2\int_{0}^1 t-t^{x} dt=2\left(\frac{1}{2}-\frac{1}{x+1}\right)=1-\frac{2}{x+1} $$ And now we can determine $x$ as you did.

Answer 2

Let's prove by induction that $$\sum_{m=2}^n \frac1{\sum_{k=1}^m k} = \frac{n-1}{n+1}$$ The base case $n=2$ is true. Now the induction step: $$\sum_{m=2}^{n+1} \frac1{\sum_{k=1}^m k}=\sum_{m=2}^n \frac1{\sum_{k=1}^m k}+\frac1{\sum_{k=1}^{n+1}k}=\frac{n-1}{n+1}+\frac2{(n+1)(n+2)}=\frac{n^2+n}{(n+1)(n+2)}=\frac{n}{n+2}$$

Therefore the answer is $n=2012$.

(I know this is kind of lame, but the formula is easy to guess)

Answer 3

I'm not sure there is much which can be done radically differently. The proof can be recast as an induction - which is what you might do after computing the first few sums by hand as $\frac 13, \frac 12=\frac 24, \frac 35, \frac 23=\frac 46, \frac 57$

The $n^{th}$ term is $\cfrac 2{(n+1)(n+2)}$

We can prove that the sum of the first $n$ terms is $\cfrac n{n+2}$, since this is true for the first term and $$\frac n{n+2}+\frac 2{(n+2)(n+3)}=\frac {n+1}{n+3}$$

Answer 4

My approach is as follows:

$$\dfrac{2011}{2013}-\dfrac{1}{3}=\dfrac{1340}{2013}$$

$$\dfrac{1340}{2013}-\dfrac{1}{6}=\dfrac{2009}{4026}$$

$$\vdots$$

$$\dfrac{4}{4048143}-\dfrac{1}{2023066}=\dfrac{1}{2025078}$$

$$\dfrac{1}{2025078}-\dfrac{1}{2025078}=0$$

So $x=2012$.