How Can the Ratio Test Be Proven for Series Divergence When the Ratio Exceeds 1?

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The discussion focuses on proving the ratio test for series divergence when the ratio exceeds 1. Participants suggest comparing the series in question to a geometric series as a starting point. An example is provided where if the ratio is greater than 1, one can demonstrate that the series terms will eventually exceed any finite value by considering the addition of a nonzero number infinitely. This approach highlights the fundamental concept that terms growing larger than a converging series will lead to divergence. The conversation emphasizes the need for a clear understanding of series behavior in relation to the ratio test.
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Homework Statement
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Relevant Equations
Please see below
For (a) and (b),
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Does someone please know how to prove this? I don't have any ideas where to start.

Thanks!
 
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a) b) is standard theory.

Relevant examples also included in the link above.
 
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A good starting point is to compare it to a geometric series. For example if ##c=1/3## can you think of a series that converges whose terms are eventually guaranteed to be larger than the ##x_n##?
 
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If the ratio is > 1, then compare to adding a nonzero number to itself " infinitely often", show it will eventually surpass any finite value.
 
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First, I tried to show that ##f_n## converges uniformly on ##[0,2\pi]##, which is true since ##f_n \rightarrow 0## for ##n \rightarrow \infty## and ##\sigma_n=\mathrm{sup}\left| \frac{\sin\left(\frac{n^2}{n+\frac 15}x\right)}{n^{x^2-3x+3}} \right| \leq \frac{1}{|n^{x^2-3x+3}|} \leq \frac{1}{n^{\frac 34}}\rightarrow 0##. I can't use neither Leibnitz's test nor Abel's test. For Dirichlet's test I would need to show, that ##\sin\left(\frac{n^2}{n+\frac 15}x \right)## has partialy bounded sums...