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We know that the sequence $$1+\frac12+\cdots+\frac1n$$ diverges but what if, we just multiply and divide every term by $n$ to get $$\left(n+\frac n2+\cdots+ \frac n n\right)/n$$ and now if we apply the cauchy's first theorem on limit of sequence which will give the limit of seq to be $1.$ Giving a contradiction

Now I wonder How is that even possible, I think that we cannot really take terms of the sequence depending on $n.$

Help !! Thnx in advance

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    $\begingroup$ Hi, can you please explain what do you mean Cauchy's first theorem? (A link to wiki or similar would be enough.) I don't seem to be able to google what exactly it is and it seems to be significant. $\endgroup$
    – yo'
    Dec 25, 2017 at 19:08
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    $\begingroup$ You can't "divide every term by $n$" since each term is a function of $n$. You have just rewritten the $n$th partial sum, without changing its value. So you haven't caused the sequence of partial sums to converge. $\endgroup$ Dec 25, 2017 at 19:10
  • $\begingroup$ @yo I could help you with this priti2212.blogspot.in/2013/05/… $\endgroup$ Dec 25, 2017 at 19:17
  • $\begingroup$ @EthanBolker I have not only divided , i have multiplied also isn't it makes sense $\endgroup$ Dec 25, 2017 at 19:21
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    $\begingroup$ I take back my words. The answer of @EthanBolker illustrates it better. $\endgroup$
    – Shashi
    Dec 25, 2017 at 19:40

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Assuming you want to apply your theorem, what are the terms of your sequence? You cannot have $x_1 = n, x_2=\frac{n}{2}, \ldots, x_n=\frac{n}{n}$ for any arbitrary $n$. You assumed that the terms form a sequence, but they do not, as it is not coherent. What I mean is: every time you write $\frac{\frac{n}{1}+\cdots+1}{n}$, you are talking about a different $n$, so you cannot organize the terms in the denominator as a sequence. I mean, otherwise once you would have $x_{1}=m$, and once $x_{1}=m$. This is not the average of terms of a fixed sequence. The first term should not be dependent on $n$.

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    $\begingroup$ Stupid me, my comment to the question does not make sense at all! $\endgroup$
    – yo'
    Dec 25, 2017 at 19:11
  • $\begingroup$ @ElieLouia So basically My first n terms should be independent of n otherwise I cannot apply the theorem right? $\endgroup$ Dec 25, 2017 at 19:38
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    $\begingroup$ The $n$th term is surely dependent of $n$. But the first term cannot be dependent on $n$. As in, if you change the symbol $n$ by $m$, you should not get a different value. $\endgroup$
    – Higurashi
    Dec 25, 2017 at 19:40

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