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Question Number 146722 by SOMEDAVONG last updated on 15/Jul/21

$$\mathrm{D}=\underset{\mathrm{n}\to +\propto }{\lim }\frac{{\mathrm{n}}^2}{\mathrm{n}-1}[\frac{\sin \frac{88}{\mathrm{n}}}{1+2}+\frac{\sin \frac{88}{\mathrm{n}}}{1+2+3}+...+\frac{\sin \frac{88}{\mathrm{n}}}{1+2+3+...+\mathrm{n}}]$$

Answered by Olaf_Thorendsen last updated on 15/Jul/21

$${\mathrm{S}}_n=\frac{n^2}{n-1}\left[\underset{k=2}{\sum ^n}\frac{\sin \frac{88}{n}}{\underset{p=1}{\sum ^k}p}\right]$$$${\mathrm{S}}_n=\frac{n^2}{n-1}\left[\underset{k=2}{\sum ^n}\frac{\sin \frac{88}{n}}{\frac{k(k+1)}{2}}\right]$$$${\mathrm{S}}_n=\frac{n^2}{n-1}\sin \frac{88}{n}\left(\underset{k=2}{\sum ^n}\frac{2}{k(k+1)}\right)$$$${\mathrm{S}}_n=\frac{2n^2}{n-1}\sin \frac{88}{n}(\underset{k=2}{\sum ^n}\frac{1}{k}-\frac{1}{k+1})$$$${\mathrm{S}}_n=\frac{2n^2}{n-1}\sin \frac{88}{n}(\frac{1}{2}-\frac{1}{n+1})$$$${\mathrm{S}}_n=\frac{2n^2}{n-1}\sin \frac{88}{n}\left(\frac{n-1}{2(n+1)}\right)$$$${\mathrm{S}}_n=\frac{n^2}{n+1}\sin \frac{88}{n}$$$${\mathrm{S}}_n\underset{\mathrm{\infty }}{\sim }\frac{n^2}{n+1}\times \frac{88}{n}=\frac{88n}{n+1}\underset{\mathrm{\infty }}{\to }88$$

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