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By 'easy' I mean avoiding heavy notation. Ideally we only need numbers and multiplication signs. You can use algebra implicitly, but try to keep things elementary school level.

Two examples of what I mean:

The sum of reciprocals is infinite: The first term is $1$. Take the next 9 terms. The smallest one is $\frac{1}{10}$ so their sum is at least $\frac{9}{10}$. That makes 10 terms. Take the next 90 terms. The smallest one is $\frac{1}{100}$ so their sum is at least $\frac{90}{100} = \frac{9}{10}$. That makes 100 terms. Take the next 900 terms. The smallest one is $\frac{900}{1000} = \frac{9}{10}$. Proceeding in this manner, we see the sum is greater than $1+\frac{9}{10}+\frac{9}{10}+\ldots +\frac{9}{10}$ for any number of $+\frac{9}{10}$s.

The sum of powers of halves is finite: Eat half a cookie. Now eat half of whats left. Now eat half of what's left. Now eat half of what's left. The remainder halves every time. So after eating forever you eat the whole cookie. So $\frac12+\frac14+\frac18+ \ldots = 1$.

Is there a similar trick for the series $\sum_{n=1}^\infty \frac{1}{n^2}$?

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    $\begingroup$ $\frac{1}{n^2} < \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$ for $n > 1$? (That goes back at least to Bernoulli [don't remember which one].) $\endgroup$ Commented 10 hours ago
  • $\begingroup$ @DermotCraddock Give the comment as an answer. Because it is (the) one :-) $\endgroup$ Commented 10 hours ago
  • $\begingroup$ Related: Proving $\frac{1}{n^2}$ infinite series converges without integral test $\endgroup$ Commented 10 hours ago
  • $\begingroup$ At an elementary school level, one doesn't have the notion of convergence, so the question is a non-starter. $\endgroup$ Commented 1 hour ago
  • $\begingroup$ @JyrkiLahtonen and other close-voters: This question already explains pretty clearly how it’s different from the proposed dupe questions, in the level of exposition it’s after; and looking at their (many) answers, most clearly don’t fit what this question is after The only answer that really fits here is Jyrki’s answer by picture, which is beautiful and answers this very well, possibly with a little expansion. $\endgroup$ Commented 55 mins ago

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\begin{align*} \frac11+\frac14+\frac19+\dots &\leq \frac{1}{1}+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}\dots \\ &= \frac11+\frac11-\frac12+\frac12-\frac13+\frac13-\frac14\dots\\ &= 2 \end{align*}

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    $\begingroup$ That is one of the solutions to the suggested duplicate target, also here. $\endgroup$ Commented 7 hours ago

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