Question Number 5753 by Rasheed Soomro last updated on 26/May/16
$$\mathrm{If}\:\:\mathrm{0}<{r}<\mathrm{1},\:\mathrm{determine} \\ $$$$\mathrm{1}+\frac{{ar}}{{a}+{r}}+\frac{{ar}^{\mathrm{2}} }{{a}+\mathrm{2}{r}}+\frac{{ar}^{\mathrm{3}} }{{a}+\mathrm{3}{r}}….. \\ $$
Commented by FilupSmith last updated on 26/May/16
$${S}=\mathrm{1}+{T} \\ $$$${T}=\frac{{ar}}{{a}+{r}}+\frac{{ar}^{\mathrm{2}} }{{a}+\mathrm{2}{r}}+… \\ $$$${T}={ar}\left(\frac{\mathrm{1}}{{a}+{r}}+\frac{{r}}{{a}+\mathrm{2}{r}}+…\right) \\ $$$${T}=\frac{{ar}}{{a}+{r}}+{ar}\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{r}^{{i}} }{{a}+\left({i}+\mathrm{1}\right){r}}\right) \\ $$$$\therefore{S}=\mathrm{1}+\frac{{ar}}{{a}+{r}}+{ar}\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{r}^{{i}} }{{a}+\left({i}+\mathrm{1}\right){r}}\right) \\ $$$${this}\:{is}\:{very}\:{complicated} \\ $$$${see}\:{further}\:{comments} \\ $$
Commented by FilupSmith last updated on 26/May/16
Commented by FilupSmith last updated on 26/May/16
Commented by Rasheed Soomro last updated on 26/May/16
$${T}=\frac{{ar}}{{a}+{r}}+{ar}\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{r}^{{i}} }{{a}+\left({i}+\mathrm{1}\right){r}}\right) \\ $$$${i}=\mathrm{1}\:{or}\:{i}=\mathrm{0}? \\ $$$${T}=\frac{{ar}}{{a}+{r}}+{ar}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{r}^{{i}} }{{a}+\left({i}+\mathrm{1}\right){r}}\right) \\ $$
Commented by FilupSmith last updated on 27/May/16
$$\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{ar}^{{i}+\mathrm{1}} }{{a}+\left({i}+\mathrm{1}\right){r}}\right)=\frac{{ar}}{{a}+{r}}+{ar}\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{r}^{{i}} }{{a}+\left({i}+\mathrm{1}\right){r}}\right) \\ $$
Commented by Rasheed Soomro last updated on 27/May/16
$${Ok}\:{than}\Bbbk{s}! \\ $$