There is an MSE post stating the following question.
I pick a number n from 1 to 100. If you guess correctly, I pay you $n and zero otherwise. How much would you pay to play this game?
The answer given to question above is $$\frac{1}{\sum_{k=1}^{100} \frac{1}{k}}.$$
Let us consider a slightly difference game. If I pay you $n^2$ instead, how much would you pay to play this game?
Intuitively, we should pay more now as we are getting higher reward.
I guess we have $$i^2 p_i = j^2 p _j \quad \text{for all }1\leq i,j\leq 100$$ where $p_i$ is the probability of choosing number $i.$ The equations above are obtained from indifference criterion.
On the other hand, we also have $$\sum_{k=1}^{100} p_k = 1,$$ which implies that $$\sum_{k=1}^{100}\frac{1}{k^2} p_1 = 1$$
So, the expected payoff is $$p_1 = \frac{1}{\sum_{k=1}^{100} \frac{1}{k^2}}.$$ Is my reasoning above correct?
Yes, you are right. Note that this is approximately
$$ \frac1{\sum_{k=1}^\infty\frac1{k^2}}=\frac6\pi\;. $$
Thus, interestingly, since the sum over $\frac1{k^2}$ converges whereas the sum over $\frac1k$ diverges, this game has approximately the same value independent of $n$ (for sufficiently large $n$), whereas the value of the original game tends to $0$ with $n\to\infty$.