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Show that $\displaystyle\lim_{n\to \infty} (n-1)^k-n^k=0$ for $0 < k <1$.

Use the fact that: $0 \le (n-1)^k-n^k = n^k((1-\frac 1 {n})^k-1)$

I'm staring at this problem clueless for some time now, I don't see how squeezing would work, using the limit definition with epsilon get me to a dead end as well... Any hints please?

Note: no Taylor, LHR, integrals or derivatives.

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  • $\begingroup$ Can you use l'Hospital's rule? $\endgroup$
    – Timbuc
    Commented Dec 2, 2014 at 22:01
  • $\begingroup$ how do you define $n^k$ for $0<k<1$? $\endgroup$
    – Lukas
    Commented Dec 2, 2014 at 22:01
  • $\begingroup$ Well @kuhaku, then it is going to be, apparently, one hell of calculations, as you can use almost nothing that could, probably, make this thing "softer". For one, I've no idea how to attack it just like that. Good luck, it's an interesting question. $\endgroup$
    – Timbuc
    Commented Dec 2, 2014 at 22:03
  • $\begingroup$ @Lukas I think we define it as the kth root of n. $\endgroup$
    – shinzou
    Commented Dec 2, 2014 at 22:04
  • $\begingroup$ well actually, to define $n^k$ you have to use some sort of infinite series expansion, at least I can't think of another way. $\endgroup$
    – Lukas
    Commented Dec 2, 2014 at 22:07

1 Answer 1

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You have $1 > (1-\frac{1}{n})^{k} > (1-1/n)$ so that $|1-(1-\frac{1}{n})^{k}| \leqslant \frac{1}{n}$ and $|n^{k}(1-(1-\frac{1}{n})^{k})| \leqslant |n^{k-1}|$.

But indeed $n^{k-1}\rightarrow 0$ so you can apply your squeezing theorem, since, as pointed out in the question, $(n-1)^{k}-n^{k}=n^{k}((1-1/n)^{k}-1)$.

I don't know where you got this problem but I find it to be a really good one. Taylor, integrals, derivatives, all ways of killing flies with atomic bombs.

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  • $\begingroup$ So if you multiplied this $|1-(1-\frac{1}{n})^{k}| \leqslant \frac{1}{n}$ by $n^k$, where did the $1$ go? $\endgroup$
    – shinzou
    Commented Dec 2, 2014 at 22:44
  • $\begingroup$ Can you please explain the process for getting the last inequality? Let $w=(1 - 1/n )^k$. Then we have $|1-w|< 1/n$. From here , how did you get : $|w|< 1/n$ ? $\endgroup$
    – Srinivas K
    Commented Dec 2, 2014 at 22:50
  • $\begingroup$ @Srinivas K I'm sorry I forgot to write down the $1$ in my post, I'll correct that. $\endgroup$
    – Sergio
    Commented Dec 2, 2014 at 23:11
  • $\begingroup$ I think I'm missing some basic point here. Can you please explain the next step for applying the squeeze theorem ? Is this correct :$n^k (( 1 - 1/n )^k - 1) < |n^k (1-( 1 - 1/n )^k )|< |n^{k-1}| -> 0 $ ? $\endgroup$
    – Srinivas K
    Commented Dec 2, 2014 at 23:29
  • $\begingroup$ @Srinivas K What we proved is that $|(n-1)^{k}-n^{k}|=|n^{k}(1-(1-\frac{1}{n})^{k})|<|n^{k-1}|$. So $0<|(n-1)^{k}-n^{k}|<|n^{k-1}|$, by squeeze theorem $|(n-1)^{k}-n^{k}|\rightarrow 0$, which implies that $(n-1)^{k}-n^{k}\rightarrow 0$. $\endgroup$
    – Sergio
    Commented Dec 2, 2014 at 23:54

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