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10

It converges to $0$, in fact $\frac{1}{k^7\sin{k}}$ already converges to $0$, see Theorem 2 here. This theorem gives a nice characterization of the irrationality measure of $\pi$ as the borderline number $\mu$ such that $\frac{1}{k^{u-1}\sin{k}}$ converges to $0$ for $u>\mu$, and diverges for $u<\mu$. So $\frac{1}{k^7\sin{k}}$ converges because $\mu$ ...

0

NOT A PROOF. But here are my thoughts. I don't think it will converge. As k continues to get larger, it will hit closer and closer multiples of pi. For example, k=3, 31, 314, 3141, 31415, ... etc. As these values of k occur, sin(k) will get closer and closer to zero. Since sin(k) is in the denominator, the effect will cause the $k^{th}$ sequence element to ...

0

I believe Bakers theorem can put a bound on the number of solutions to this equation.

0

Assuming $a,b$ are integers, it can be verified by brute-force search that if there are any solutions, then $a\ge4000$ or $b\ge2000$. Checked with Mathematica with following code: Monitor[Do[If[Abs[Log[5]/Log[7] - a/b] <= 1/7^b, Print[{a, b}]], {b, 1, 1999}, {a, 1, 3999}], {b, a}] Using Will Jagy's advice, checking continued fractions up to 19 terms ...

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