proof hint:pic.twitter.com/BhjKswj9c5
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you can't sum all the natural numbers, but you can sum the reciprocal of the first 𝑛pic.twitter.com/u46FdR83qx
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What are the assumptions here?
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What do you mean?
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1/(1+...+n) = 2/[n(n+1)] = 2/n - 2/(n+1), so the sum reduces to the telescoping sum: 2[\Sum_2^n 1/n - 1/(n+1)] = 2(1/2 -1/3 + 1/3 -1/4 + ...) = 2/2 = 1.
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Oooh that’s interesting! Let me see if I can figure out how that works
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1+2+...+n=1/2 n(n+1) So 1/(1+2+...+n)=2/[n(n+1)] So what we have is 2/[2*3]+2/[3*4]+2/[4*5]+...
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Not a believer in -1/12?
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I thought I would understand this, but I don't ;-;
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At first I was getting ans=2, then I realized you sneaked an index thinger in there!
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