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# HELP!!

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Find the sum of the infinite series $$1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots$$.

May 17, 2020

#1
-1

You can write this as$$\frac{1}{1-\frac{1}{1998}}+\frac{\frac{1}{1998}}{1-\frac{1}{1998}}...$$

That can be written as $$\frac{\frac{1}{1-\frac{1}{1998}}}{{1-\frac{1}{1998}}}$$

'

Hope this helps

May 18, 2020
#2
+25543
+2

Find the sum of the infinite series

$$1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots$$

$$\small{ \begin{array}{rclllllllllc} & & \mathbf{1}&\mathbf{+} & \mathbf{2\left(\dfrac{1}{1998}\right)} &\mathbf{+} & \mathbf{3\left(\dfrac{1}{1998}\right)^2} &\mathbf{+} &\mathbf{ 4\left(\dfrac{1}{1998}\right)^3} & \mathbf{+}&\mathbf{\cdots} \\\\ & & & & & & & & & & &\mathbf{\text{sum }=\dfrac{a}{1-r}} \\ &=& 1&+ & 1\left(\dfrac{1}{1998}\right)^1 &+ & 1\left(\dfrac{1}{1998}\right)^2 &+ & 1\left(\dfrac{1}{1998}\right)^3 &+&\cdots & \dfrac{1}{1-\dfrac{1}{1998}} \\\\ & & &+ & 1\left(\dfrac{1}{1998}\right)^1 &+ & 1\left(\dfrac{1}{1998}\right)^2 &+ & 1\left(\dfrac{1}{1998}\right)^3 &+&\cdots & \dfrac{\left(\dfrac{1}{1998}\right)^1}{1-\dfrac{1}{1998}} \\\\ & & & & &+ & 1\left(\dfrac{1}{1998}\right)^2 &+ & 1\left(\dfrac{1}{1998}\right)^3 &+&\cdots & \dfrac{\left(\dfrac{1}{1998}\right)^2}{1-\dfrac{1}{1998}} \\\\ & & & & & & & + & 1\left(\dfrac{1}{1998}\right)^3 &+&\cdots & \dfrac{\left(\dfrac{1}{1998}\right)^3}{1-\dfrac{1}{1998}} \\\\ & & & & & & & & & +&\ldots \\ \end{array} }$$

$$\begin{array}{|rcll|} \hline &&\mathbf{ 1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots } \\\\ &=& \dfrac{1}{1-\dfrac{1}{1998}} + \dfrac{\left(\dfrac{1}{1998}\right)^1}{1-\dfrac{1}{1998}} + \dfrac{\left(\dfrac{1}{1998}\right)^2}{1-\dfrac{1}{1998}} + \dfrac{\left(\dfrac{1}{1998}\right)^3}{1-\dfrac{1}{1998}} +\cdots \\\\ &=& \dfrac{1}{1-\dfrac{1}{1998}} \left( 1+ \left(\dfrac{1}{1998}\right)^1+\left(\dfrac{1}{1998}\right)^2+\left(\dfrac{1}{1998}\right)^3 +\cdots \right) \\\\ &=& \dfrac{1}{\dfrac{1998-1}{1998}} \left( 1+ \left(\dfrac{1}{1998}\right)^1+\left(\dfrac{1}{1998}\right)^2+\left(\dfrac{1}{1998}\right)^3 +\cdots \right) \\\\ &=& \dfrac{1998}{1997} \left( 1+ \left(\dfrac{1}{1998}\right)^1+\left(\dfrac{1}{1998}\right)^2+\left(\dfrac{1}{1998}\right)^3 +\cdots \right) \\\\ &=& \dfrac{1998}{1997} \left( \dfrac{1}{1-\dfrac{1}{1998}} \right) \\\\ &=& \dfrac{1998}{1997} \left( \dfrac{1}{\dfrac{1998-1}{1998}} \right) \\\\ &=& \dfrac{1998}{1997}\times \dfrac{1998}{1997} \\\\ &=& \dfrac{1998^2}{1997^2} \\\\ &=&\mathbf{ \dfrac{3992004}{3988009}} \\ \hline \end{array}$$

May 18, 2020
#3
+25543
+2

Find the sum of the infinite series
$$1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots$$.

$$\text{Let r=\dfrac{1}{1998} }$$

$$\begin{array}{|rcll|} \hline s&=& \mathbf{1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots} \\ s&=& \mathbf{1+2r +3r^2+4r^3+\cdots + nr^{n-1} + \ldots} \\ \hline \end{array}$$

$$\begin{array}{|rcrl|} \hline s_n &=& 1+&2r +3r^2+4r^3+\cdots + nr^{n-1} \quad |\quad \cdot r \\ rs_n &=& & 1r +2r^2+3r^3+\cdots + (n-1)r^{n-1}+nr^n \\ \hline s_n-rs_n &=& 1+&r+r^2+r^3+\ldots + r^{n-1} -nr^n \\ \hline \end{array}\\ \begin{array}{|rclrcrl|} \hline s_n(1-r) &=& \underbrace{1+r+r^2+r^3+\ldots + r^{n-1}}_{=S_n(\text{ geometric progression})} -nr^n \\ s_n(1-r) &=& S_n -nr^n & S_n &=& 1+&r+r^2+r^3+\ldots + r^{n-1} \quad | \quad \cdot r\\ & & & rS_n &=& &r+r^2+r^3+\ldots + r^{n-1}+r^n \\ \hline & & & S_n-rS_n &=& 1- &r^n \\ & & & S_n(1-r) &=& 1- &r^n \\ & & & S_n &=& \dfrac{1- r^n}{1-r} \\ s_n(1-r) &=& \dfrac{1- r^n}{1-r} -nr^n \\ s_n &=& \dfrac{1- r^n}{(1-r)^2} - \dfrac{nr^n}{1-r} \\ s &=& \lim \limits_{n\to \infty} s_n \\ s &=& \lim \limits_{n\to \infty} \dfrac{1- r^n}{(1-r)^2} - \dfrac{nr^n}{1-r} \\ && \boxed{\lim \limits_{n\to \infty} r^n = \lim \limits_{n\to \infty} \left(\dfrac{1}{1998}\right)^n = 0} \\ s &=& \dfrac{1- 0}{(1-r)^2} - 0 \\ s &=& \dfrac{1}{(1-r)^2} \\ s &=& \dfrac{1}{\left(1-\dfrac{1}{1998}\right)^2} \\ s &=& \left(\dfrac{1998}{1997}\right)^2 \\ \\ s &=& \dfrac{1998^2}{1997^2} \\\\ s &=&\mathbf{ \dfrac{3992004}{3988009}} \\\\ \mathbf{s} &=& \mathbf{1.00100175300507095\ldots...} \\ \hline \end{array}$$

May 18, 2020
edited by heureka  May 18, 2020
#4
-1

I tried this and it was wrong??

May 18, 2020
#5
+25543
+2

Find the sum of the infinite series $$1+2\left(\dfrac{1}{1998}\right)+3\left(\dfrac{1}{1998}\right)^2+4\left(\dfrac{1}{1998}\right)^3+\cdots$$

heureka  May 18, 2020