Isn’t this already the result of your 1st formula? As the denominator of the last fraction you wrote down, (n+1)/n, cancels out with the counter of the one right before, n/(n-1), which you didn’t write down. Thus the whole product up to the nth term reads after cancellation of neighbouring counters and denominator pairs (n+1)/1 →∞ when n→∞.
Yes - I mostly left the first part in for the humor (it was legitimately my first stab at the problem), but it gives the same result, just in a way that’s a little harder (to me, at least) to see. The cancellation is unbalanced: Each numerator cancels with the next denominator, which necessarily brings with it the next numerator - you’ve always got the next numerator in line, as-is, after any number of canceled pairs. So while everything cancels out in the limit, the product up to n equals n+1, and so the limit of the product is ∞.
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solution
Isn’t this already the result of your 1st formula? As the denominator of the last fraction you wrote down,
(n+1)/n, cancels out with the counter of the one right before,n/(n-1), which you didn’t write down. Thus the whole product up to the nth term reads after cancellation of neighbouring counters and denominator pairs(n+1)/1 →∞ when n→∞.reply
Yes - I mostly left the first part in for the humor (it was legitimately my first stab at the problem), but it gives the same result, just in a way that’s a little harder (to me, at least) to see. The cancellation is unbalanced: Each numerator cancels with the next denominator, which necessarily brings with it the next numerator - you’ve always got the next numerator in line, as-is, after any number of canceled pairs. So while everything cancels out in the limit, the product up to n equals n+1, and so the limit of the product is ∞.