Let $a$ be a series of natural numbers where
$$
a_k=\begin{cases}
\frac{2(k+1)}{3} & \text{if }(k \mod3)=2\\
1 & \text{otherwise}
\end{cases}
$$
for every $k\in\mathbb{Z}^+$. The question is what is the average of the reciprocal of all these elements. Expressed as a summation it looks like this:
$$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{a_k}
$$
Now if we look at the series, it looks like this $\{1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, 1, 1, 14, 1, 1, 16, 1, \dots\}$. What I tried to do was split the summation into three summations, becuase notice that there are groups of three. First, comes a $1$, then comes that value from the mod and then another $1$. I tried grouping these in the summation like this:
$$\lim_{n\to\infty}\frac{1}{n}\left(\left(2\sum_{k=1}^{n/3}1\right)+\sum_{m=1}^{n/3}\frac{1}{\frac{2(3(m-1)+1)}{3}}\right)=\lim_{n\to\infty}\frac{1}{n}\left(\frac{2n}{3}+\sum_{m=1}^{n/3}\frac{1}{2m}\right)
$$
I tried various methods to solve it from here but I kept getting $2$ as a result, which should be impossible $a$ is a series for natural numbers and the highest value this should return should be $1$ for an infinite series of $1$-s.
Any help is appreciated. If you don't tell me the answer, that's also fine as long as someone tells me what I did wrong and how I should have done it or how I should solve the limit, because at this point I have no idea.
Best Answer
After looking at the thing I wrote, I realized how stupid am I. My problem was not dividing the $2n$ by 3, which lead to me getting $2$ as a result. I'll still post the solution to the sum.
The partial sum of that summation looks like this: $$\frac{2}{3}+\lim_{n\to\infty}\frac{\gamma+\psi^{(0)}(1+\frac{n}{3})}{n} $$ The limit evalutes to $0$ thus the answer is $0.\bar{6}$, which does satisfy the condition of it being lower than $1$.