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Thread: limit question (indeterminate forms??)

  1. #1
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    limit question (indeterminate forms??)

    Question:
    We work in the real numbers. Are the following true or false? Give a proof or counterexample.
    (a) If $\displaystyle \sum a^4_n$ converges, then $\displaystyle \sum a^5_n$ converges.
    (b) If $\displaystyle \sum a^5_n$ converges, then $\displaystyle \sum a^6_n$ converges.
    (c) If $\displaystyle a_n \geq 0$ for all $\displaystyle n$, and $\displaystyle \sum a_n$ converges, then $\displaystyle na_n \rightarrow 0$ as $\displaystyle n \rightarrow \infty$.
    (d) If $\displaystyle a_n \geq 0$, for all $\displaystyle n$, and $\displaystyle \sum a_n$ converges, then $\displaystyle n(a_n - a_{n-1}) \rightarrow 0$ as $\displaystyle n \rightarrow \infty$.
    (e) If $\displaystyle a_n$ is a decreasing sequence of positive numbers, and $\displaystyle \sum a_n$ converges, then $\displaystyle na_n \rightarrow 0$ as $\displaystyle n \rightarrow \infty$.

    MY WORK:
    (a) and (b) can be proved similarly. Since $\displaystyle \sum a_n^4$ converges, for some $\displaystyle N$, when $\displaystyle n \geq N$, then $\displaystyle a_n^4 < 1$. Take $\displaystyle \beta$ s.t., $\displaystyle a_n^4 < \beta < 1$. That is, $\displaystyle a_n < (\beta)^{1/4} < 1$. Also, $\displaystyle |(\beta)^{1/5}| < 1$. This implies that $\displaystyle |a_n^5| < 1$ and therefore $\displaystyle \sum |a_n^5|$ converges.

    (c) This is where I get confused. THis seems like an indeterminate form, we have never done this in class. same for (d) and (e). any suggestions?
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  2. #2
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    Your work on part a) could a bit smoother: $\displaystyle (a_n)^4<1$ implies
    $\displaystyle |a_n|^5\le (a_n)^4<1$.

    For part b) consider $\displaystyle a_n = \dfrac{{\left( { - 1} \right)^n }}{{\sqrt[{10}]{n}}}$.
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  3. #3
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    Here is part e).
    For ease of notation use $\displaystyle S_N = \sum\limits_{k = 1}^N {a_k } $ for the partial sums.
    Because the series converges , the sequence $\displaystyle (S_n)$ is Cauchy.
    If $\displaystyle \varepsilon > 0$ there is positive integer $\displaystyle K$ such if $\displaystyle j>K$ then $\displaystyle S_j-S_K<\frac{\varepsilon }{2}$.
    We also know that $\displaystyle (a_n)\to 0$. So there is a positive integer $\displaystyle J>K$ such that $\displaystyle n \geqslant J\, \Rightarrow \,a_n < \frac{\varepsilon }{{2K}}$.

    Now we have setup the machinery, we just have to find the tricks.
    If $\displaystyle p\ge J$ then $\displaystyle pa_p=(p-K)a_p+Ka_p$.
    But we notice that $\displaystyle (p-K)a_p\le S_p-S_K.$

    Can you finish?
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