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Again, shattered by this question on series, I did have no clue how to begin. Sequences limits are approached through absolute values of the $n$-th term and the assumed limit being smaller than a given delta. And this for a given $N$. I don't see the link with power-series exercises, as for an example, the following:

Find a $K$, such that for all $n \geq K$ we have $$\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \ldots + \frac{1}{\sqrt{n}} > 1000$$

Help me with this one to, but more interesting is, how the theory on sequences applies to these problems. Or don't they?

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  • $\begingroup$ the epsilon delta argument for divergence is slightly different than that for convergence, in order to prove that a sequence diverges, we say that for any epsilon, there exists a delta s.t. n>delta implies S(n) > epsilon. An easy and surefire delta for that series is is delta = epsilon^2, for the minimum delta a slightly more complex formula would be necessary. This problem above is just giving a example epsilon (1000) and asking for any specific delta. $\endgroup$ – Math chiller Aug 14 '16 at 5:17
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You are not asked for the minimum $K$, but just for one that works. You want $\sum_{i=1}^K \frac 1{\sqrt i} \gt 1000$. One way to get there is notice that all the terms are at least $\frac 1 {\sqrt K}$ and there are $K$ of them. So the sum is at least $K\cdot \frac 1 {\sqrt K}= \sqrt K$. $K=1000^2$ therefore works.

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A quite good bound can be found by comparing with an integral. Since $f(x)$ is decreasing and integrable on the positive real axis we have that $\sum_{k=1}^Nf(k)\geq\int_{1}^{N+1}f(x)dx=2\sqrt{N+1}-2$ and $2\sqrt{N+1}-2\geq 1000$ happens when $N\geq 501^2-1$

Added 1: Given that $N=250732$ is the smallest possible $N$, we see that this bound is $251000-250732=268$ larger than the minimal $N$.

Added 2: Just realized that another (maybe an even nicer) approach to obtain the bound above, is to note that $\frac{1}{\sqrt{k}}\geq{}\frac{2}{\sqrt{k+1}+\sqrt{k}}=2(\sqrt{k+1}-\sqrt{k})$, which implies that $\sum_{k=1}^Nf(k)\geq \sum_{k=1}^N2(\sqrt{k+1}-\sqrt{k})=2\sqrt{N+1}-2$.

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Hint: Use ideas from integral estimation of series to get a good estimate. If $f(x)$ is a continuous positive decreasing function and $a_n =f(n)$ then $$ a_n + \int_1^n f(x)dx \le a_1+a_2+\cdots+a_n \le a_1+\int_1^nf(x)dx$$ Draw the graph of $f(x)$ and compare area under it over $[1,n]$ with areas of reactangels built on left or right end points of $[i,i+1]$; as you do for Riemann integral estimation. Then use your integral to get an estimate for $K$.

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The series you wrote is 'greater' (term by term) than the harmonic series, therefore it diverges, and that means it can get as big as you want it to as $n$ is large enough. To find $K$ such that for $n \geq K$ your series will be greater than $1000$ seems more like a programming question. You cannot calculate by hand all the terms, so try and make a program which calculates the sum, and then you will find the value of $K$. The mathematical part of the problem tells you that such a $K$ does exist.

After a quick try it seems that the smallest $K$ is $250732$

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  • $\begingroup$ I would argue that to find such a $K$ does not have to be a "programming problem", and that one of the principal purposes of real analysis is to solve just this sort of problem without requiring a brute-force calculation. But I think the higher-scoring answers in this section have made that argument more eloquently than I could. $\endgroup$ – MJD Apr 16 '12 at 17:50
  • $\begingroup$ @MarkDominus: I assure you that this is not one of the principal purposes of real analysis. And I don't think that pointing out that the only way to find the best possible constant $K$ in a reasonable amount of time is by writing a small program. I know that the problem asks for a $K$. I gave the best $K$. $\endgroup$ – Beni Bogosel Apr 16 '12 at 18:58
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The core of the $\epsilon - \delta$ limit theory of sequences is that $\forall\epsilon>0\exists\delta: x<\delta\Leftrightarrow|L - f(\delta)| < \epsilon$. In the case of divergence we have $L=\infty $ which gives us $L-\infty = \infty > \epsilon \forall\epsilon\in\mathbb{R}$. So the $\epsilon - \delta$ argument instead becomes $\forall\epsilon>0\exists\delta: n>\delta \Leftrightarrow f(n) > \epsilon$. This problem relates to that arguent by giving the specific $\epsilon=K=1000$ and asking for a delta.

The way $\epsilon - \delta$ proofs are formulated is by expressing delat as a function of epsilon, in this case a simple and obvious choice (as stated above by @ross) is $\delta=\epsilon^2\Rightarrow \sum_{i=1}^{n\geq\delta}\frac{1}{\sqrt{i}}>\epsilon^2*\frac{1}{\sqrt{\epsilon^2}}=\epsilon$ and so in this case 1,000,000 is a good choice of delta.

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