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The partial sum $$s_n=\frac{1}{4}(3^n\sin\frac{a}{3^n}-\sin\ a)$$

$$=>S=\lim_{n\to\infty} s_n=\frac{1}{4}\lim_{n\to\infty}\left[a\frac{\sin\frac{a}{3^n}}{\frac{a}{3^n}}-\sin \ a\right]\ \ (*)$$

My question is about how that change occurred in (*). How it went from $3^n\sin\frac{a}{3^n}$ to $a\frac{\sin\frac{a}{3^n}}{\frac{a}{3^n}}$. (I know how to evaluate that limit)

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up vote 2 down vote accepted

Just sneaking in a multiplicative factor of $1$ and a bit of algebra: $$ \eqalign{ 3^n\sin\frac{a}{3^n} &=3^n\cdot{a\over a}\cdot\sin\frac{a}{3^n} \cr &={a\over a/3^n}\sin\frac{a}{3^n}\cr &=a\frac{\sin\frac{a}{3^n}}{\frac{a}{3^n}}.}$$


Incidentally, the motivation here was to use the fact that $\lim\limits_{x\rightarrow0}{\sin x\over x}=1$. The above was done to take advantage of this. We'd almost have it by writing $3^n\sin\frac{a}{3^n} ={ \sin\frac{a}{3^n}\over 1/3^n}$; but not quite, hence multiplying by $a/a$...

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you're a genius (always). Thanks a lot –  Andrew Jan 9 '12 at 17:22

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