Wayward times

Suppose that positive integer $N$ can be written as a sum of $n$ consecutive positive integers starting with $m+1$ , where $n\ge2$ and $m\ge0$ : $\begin{array}{c}\\ N &=m+1+m+2+...+m+n \\ &=mn+\frac{n(n+1)}{2} \end{array}$ Which gives $2N=n(n+2m+1) \ \star$ . RHS of $\star$ is two factors greater than $1$ with different parity, so $N$ cannot be a power of $2$ .

For $N=ab$ with $a>1$ odd and $b$ even, set $n=a$ and $m=\frac{2b-a-1}{2}$ . For odd numbers greater than $1$ , $N=2k+1$ with $k\ge1$ , one can see easily that $N=k+k+1$ . Therefore $\text{S}= \left\{ 1, 2, 2^2, 2^3, ...\right\}$ and $1+\frac{1}{2}+\frac{1}{2^2}+...=2$ .

Note that $\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2$ , which means that $1+2+\cdots+n=\dfrac{n(n+1)}2$ .

Also note that:

• $\displaystyle(m+1)+(m+2)+\cdots+n$
• $=[1+2+\cdots+n]-[1+2+\cdots+m]$
• $=\displaystyle\sum_{i=1}^ni-\sum_{i=1}^mi$
• $=\displaystyle\frac{n(n+1)}2-\frac{m(m+1)}2$
• $=\displaystyle\frac12[n(n+1)-m(m+1)]$
• $=\displaystyle\frac12[n^2+n-m^2-m]$
• $=\displaystyle\frac12[n^2-m^2+n-m]$
• $=\displaystyle\frac12[(n-m)(n+m)+(n-m)]$
• $=\displaystyle\frac12[(n-m)(n+m+1)]$

Note that exactly one of $(n-m)$ and $(n+m+1)$ is odd.

Also note that $N$ must be written as a sum of at least 2 consecutive positive integer, so $n-m\ge2$

Let $N=2^rx$ where $x$ is odd and $r$ is a non-negative integer.

We try to write $N$ in the form of $\dfrac12[(n-m)(n+m+1)]$ : $N=2^rx=\frac12(2^{r+1}x)$

Note that the second term, $(n+m+1)$ , must be bigger than the first term, $(n-m)$ .

CASE ONE:

If $2^{r+1}>x$ , then: $\begin{cases}n-m=x\\n+m+1=2^{r+1}\end{cases}$ $\implies\begin{cases}n=\frac{2^{r+1}+x-1}2=2^r+\frac{x-1}2\\m=\frac{2^{r+1}-x-1}2=2^r-\frac{x+1}2\end{cases}$

Also, $n-m\ge2$ :

• $\displaystyle\left({2^r+\frac{x-1}2}\right)-\left({2^r-\frac{x+1}2}\right)\ge2$
• $\displaystyle\frac{x-1}2+\frac{x+1}2\ge2$
• $\displaystyle x\ge2$
• which means that $x$ cannot be $1$ , which means that $N$ cannot be $2^a$ .

CASE TWO:

If $2^{r+1}<x$ , then: $\begin{cases}n-m=2^{r+1}\\n+m+1=x\end{cases}$ $\implies\begin{cases}n=\frac{2^{r+1}+x-1}2=2^r+\frac{x-1}2\\m=\frac{-2^{r+1}+x-1}2=-2^r+\frac{x-1}2\end{cases}$

Also, $n-m\ge2$ :

• $\displaystyle\left({2^r+\frac{x-1}2}\right)-\left({-2^r+\frac{x-1}2}\right)\ge2$
• $\displaystyle2^{r+1}\ge2$
• $\displaystyle r\ge0$
• which does not generate any further limitation.

THEREFORE:

The only requirement for $N$ is that it is not an integral power of $2$ , or $N\ne2^a$ .

Therefore, $S$ contains only $2^a$ .

Therefore, the required sum:

• $=\dfrac1{2^0}+\dfrac1{2^1}+\dfrac1{2^2}\cdots$
• $=\dfrac11+\dfrac12+\dfrac14+\dfrac18+\cdots$
• $=I$
• $\implies2I=\dfrac21+\dfrac22+\dfrac24+\dfrac28+\cdots$
• $\implies2I=2+\dfrac11+\dfrac12+\dfrac14+\cdots$
• $\implies2I=2+I$
• $\implies I=\fbox{2}$

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Appendix:

Its convergence can be demonstrated by the fact that the ratio between the next term to the current term is always $\dfrac12$ , which is between $-1$ and $1$ .