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I'm a student and, while playing with my calculator, find out that: $$\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}.... = \frac{1}{n-1}$$

is it a well defined equivalence and what is its name, is there a proof for that?

if we put it this way: $$1+\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}.... = \frac{n}{n-1}$$

what do you call the last term (the sum), the complementary inverse or reciprocal of 1/n?

2 Answers2

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This is a geometric series. It has a summation formula $$ \sum_{j=0}^\infty z^j = \frac{1}{1-z}, |z| < 1 $$ Plugging in $z = 1/n$ results in your formula.

Henricus V.
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    @user11374 Just the "sum" of this geometric series. – Henricus V. Mar 25 '16 at 05:26
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    The asker wanted a PROOF of this identity, not merely a statement that the identity exists. Sie did want the name, but sie also wanted the proof. – The_Sympathizer Mar 25 '16 at 06:48
  • @mike4ty4 "proof" is a vague notion because it doesn't specify what premises one starts with. Should we use the peano axioms everytime we wish to do arithmetic? Clearly not. Consequently, if one wishes to start a "proof" by referring to a well-known identity, this is a perfectly valid answer. Although the OP seems to have some more fundamental questions about fractions(?) – Andres Mejia Mar 25 '16 at 22:41
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    @AndresMejia I think the proof given in the link to wikipedia is enough. – Henricus V. Mar 25 '16 at 22:48
  • I agree with you. I think your answer alone is sufficient imo; depending on who you're answering. Either way, I don't think answers on SE are explicitly for the OP alone. – Andres Mejia Mar 25 '16 at 22:49
  • I think the info in the answers is quite enough, can you tell me: 1/n is the inverse or reciprocal of n/1? right? what do you call 1/1-n in relation to n/1? or, what do you call 1/1-n in relation to 1/n? –  Mar 26 '16 at 06:19
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assume $n>1$. Then

$$\sum_{k=0}^{K} \left(\frac{1}{n}\right)^k=1+\frac{1}{n}+...+\frac{1}{n^{K}}=L$$

Notice however, that $\frac{1}{n}\cdot L=\frac{1}{n}+..+\frac{1}{n^{K+1}}$

And then, that when you subtract them, all but the first and last term cancel: $$L-\frac{1}{n}\cdot L=1-\frac{1}{n^{K+1}}\implies L(1-\frac{1}{n})=1-\frac{1}{n^{K+1}} \implies L=\frac{1-1/(n^{K+1})}{(n-1)/(n)}=\frac{n(1-1/(n^{K+1}))}{(n-1)}$$

Now notice that as $K \to \infty$, $\frac{1}{n^{K+1}} \to 0$, so the term cancels. Thus: $$\sum_{k=0}^{\infty} \left(\frac{1}{n}\right)^k=\frac{n}{(n-1)}$$

This is a familiar trick to derive the more general geometric series equation: geometric series

Andres Mejia
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  • Is $\frac{n}{n-1}$ the complementary reciprocal of $\frac{1}{n}$ ? –  Mar 25 '16 at 06:32
  • your question is unclear to me. Are you wondering about how we divide fractions? – Andres Mejia Mar 25 '16 at 14:28
  • No, 1/n is the inverse or reciprocal of n/1? right? what do you call 1/1-n in relation to n/1? or, what do you call 1/1-n in relation to 1/n? –  Mar 26 '16 at 06:18