Is the sum of reciprocals of this concatenated sequences of triangular numbers of increasing length divergent?

39 Views Asked by At

Is the following sequence small or large? Here, large means $\displaystyle\sum_{n\in\mathbb{N}} \frac{1}{x_n}$ diverges, and small means the series converges.

Define $\Delta x_n:= x_{n+1}-x_n.$

$$x_1= 1,$$

$$\Delta x_1 = 1,$$

$$\Delta x_2 = 1, \ \Delta x_3 = 2, $$

$$\Delta x_4 = 1, \ \Delta x_5 = 2,\ \Delta x_6 = 3, $$

$$\Delta x_7 = 1, \ \Delta x_8 = 2,\ \Delta x_9 = 3, \ \Delta x_{10} = 4, $$

and so on.

So the first few terms are:

$$ 1,2,3,5,6,8,11,12,14,17,21,22,24,27,31,36,37,39,42,46,51,57,58,\ldots $$

One idea would be to estimate the sum of each row in the diagonalisation/pascal's triangle above, but I can't think how to estimate each row well enough.

1

There are 1 best solutions below

2
On

Sum of first $n$ triangular numbers is $K_n=\dfrac{n(n+1)(n+2)}{6}$.

Now, $\dfrac{1}{1+K_n}+\dfrac{1}{1+K_n+1}+\dfrac{1}{1+K_n+1+2}+\cdots+\dfrac{1}{1+K_n+1+2+\cdots+n}\leq\dfrac{n+1}{K_n}$.