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First,we define $\displaystyle I_{1}\left ( x \right )=\frac{\sin x}{x}$, then $\displaystyle \lim_{x\rightarrow 0^+}I_{1}\left ( x \right )=1$, also we have \begin{align*} I_2\left ( x \right )&=\frac{I_1\left ( x \right )-1}{x^{2}}~,~\lim_{x\rightarrow 0^+}I_2\left ( x \right )=-\frac{1}{6}\\ I_3\left ( x \right )&=\frac{I_2\left ( x \right )+\dfrac{1}{6}}{x^2}~,~\lim_{x\rightarrow 0^+}I_3\left ( x \right )=\frac{1}{120}\\ &\cdots \\ I_n\left ( x \right )&=\frac{I_{n-1}\left ( x \right )-\displaystyle \lim_{x\rightarrow 0^+}I_{n-1}\left ( x \right )}{x^{2}} \end{align*} Now we have the following questions.

(1)$I_n(x)$ is related to bernoulli number, but how to find it.

(2)Evaluate $\displaystyle \lim_{k\rightarrow +\infty }\left [ \lim_{x\rightarrow 0^{+}}I_{2k}\left ( x \right ) \right ]~,~\lim_{k\rightarrow +\infty }\left [ \lim_{x\rightarrow 0^{+}}I_{2k+1}\left ( x \right ) \right ]~,~k\in \mathbb{Z}.$

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About the second question: using the Taylor seres $$\sin\left(x\right)=\sum_{k\geq0}\frac{\left(-1\right)^{k}x^{2k+1}}{\left(2k+1\right)!} $$ we note that $$\frac{\sin\left(x\right)}{x}=\sum_{k\geq0}\frac{\left(-1\right)^{k}x^{2k}}{\left(2k+1\right)!} $$ and so $$\lim_{x\rightarrow0^{+}}I_{1}\left(x\right)=1 $$ furthermore $$\sum_{k\geq0}\frac{\left(-1\right)^{k}x^{2k}}{\left(2k+1\right)!}-1=\sum_{k\geq1}\frac{\left(-1\right)^{k}x^{2k}}{\left(2k+1\right)!} $$ and so $$I_{2}\left(x\right)=\frac{\sum_{k\geq1}\frac{\left(-1\right)^{k}x^{2k}}{\left(2k+1\right)!}}{x^{2}}=\sum_{k\geq1}\frac{\left(-1\right)^{k}x^{2\left(k-1\right)}}{\left(2k+1\right)!}\stackrel{x\rightarrow0^{+}}{\rightarrow}-\frac{1}{6} $$ and so on, hence we have $$I_{n}\left(x\right)=\sum_{k\geq n-1}\frac{\left(-1\right)^{k}x^{2\left(k-n+1\right)}}{\left(2k+1\right)!}\stackrel{x\rightarrow0^{+}}{\rightarrow}\frac{\left(-1\right)^{n-1}}{\left(2n-1\right)!} $$ then $$\lim_{n\rightarrow\infty}\left[\lim_{x\rightarrow0^{+}}I_{2n}\left(x\right)\right]=\lim_{n\rightarrow\infty}-\frac{1}{\left(4n-1\right)!}=0 $$ and $$\lim_{n\rightarrow\infty}\left[\lim_{x\rightarrow0^{+}}I_{2n+1}\left(x\right)\right]=\lim_{n\rightarrow\infty}\frac{1}{\left(4n-3\right)!}=0.$$ About the first question, Mathematica recognizes the series as an hypergeometric function $$I_{n}\left(x\right)=\sum_{k\geq n-1}\frac{\left(-1\right)^{k}x^{2\left(k-n+1\right)}}{\left(2k+1\right)!}=\frac{\left(-1\right)^{n+1}\,_{1}F_{2}\left(1;n,n+\frac{1}{2};-\frac{x^{2}}{4}\right)}{\left(2n-1\right)!}$$ but I don't know if there are some kind of relations with the Bernoulli numbers, probably would be useful more clarifications and details in the question.

Update: A possible relation can be this. We can observe that $I_{n} $ is the $(n-1)$th coefficient of the sine's Taylor series. A similar approach can be done for the Bernoulli numbers. We consider the generating function of the Bernoulli numbers $$J_{0}\left(x\right)=\frac{x}{e^{x}-1}=\sum_{k\geq0}\frac{B_{k}}{k!}x^{k} $$ so obviously $$J_{0}\left(x\right)\underset{x\rightarrow0}{\rightarrow}1 $$ so now we take $$J_{1}\left(x\right)=\frac{\frac{x}{e^{x}-1}-1}{x}=\sum_{k\geq1}\frac{B_{k}}{k!}x^{k-1} $$ and so $$J_{1}\left(x\right)\underset{x\rightarrow0}{\rightarrow}-\frac{1}{2} $$ and so on. Iterating the process we can observe that we have a recursive formula for the Bernoulli numbers $$J_{n}=\lim_{x\rightarrow0}\frac{J_{n-1}\left(x\right)-J_{n-1}}{x} =\frac{B_{n}}{n!}\Rightarrow n!J_{n}=B_{n}.$$

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