We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

Find the sum of the infinite series

(-3/7)^2+(-3/7)^3+(-3/7)^4+(-3/7)^5+...

(-3/7)^2+(-3/7)^3+(-3/7)^4+(-3/7)^5+...

Guest Feb 6, 2012

#2**+8 **

No, this sum has a finite limit:

$$\sum_{k=2}^{\inf}(\frac{-3}{7})^k=\frac{9}{70}$$

See the following figure;

In this respect it is more like the infinite series 1 + 1/2 + 1/4 + 1/8 etc. which has the finite sum of 2. In other words it is a geometric series, the sum to infinity of which is given by:

sum = a0/(1-r) where a0 is the first term [(-3/7)^{2} here] and r [= (-3/7)] is the ratio between successive terms. Because r is less than 1, successive terms get smaller, r^{∞}→0, and the sum converges to a finite value.

$${\mathtt{sum}} = {\frac{{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right)}^{{\mathtt{2}}}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right)\right)}} \Rightarrow {\mathtt{sum}} = {\mathtt{0.128\: \!571\: \!428\: \!571\: \!428\: \!6}}$$

$${\frac{{\mathtt{9}}}{{\mathtt{70}}}} = {\mathtt{0.128\: \!571\: \!428\: \!571\: \!428\: \!6}}$$

Just noticed this is a very old unanswered question!!

Alan May 18, 2014

#2**+8 **

Best Answer

No, this sum has a finite limit:

$$\sum_{k=2}^{\inf}(\frac{-3}{7})^k=\frac{9}{70}$$

See the following figure;

In this respect it is more like the infinite series 1 + 1/2 + 1/4 + 1/8 etc. which has the finite sum of 2. In other words it is a geometric series, the sum to infinity of which is given by:

sum = a0/(1-r) where a0 is the first term [(-3/7)^{2} here] and r [= (-3/7)] is the ratio between successive terms. Because r is less than 1, successive terms get smaller, r^{∞}→0, and the sum converges to a finite value.

$${\mathtt{sum}} = {\frac{{{\mathtt{\,-\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right)}^{{\mathtt{2}}}}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{3}}}{{\mathtt{7}}}}\right)\right)}} \Rightarrow {\mathtt{sum}} = {\mathtt{0.128\: \!571\: \!428\: \!571\: \!428\: \!6}}$$

$${\frac{{\mathtt{9}}}{{\mathtt{70}}}} = {\mathtt{0.128\: \!571\: \!428\: \!571\: \!428\: \!6}}$$

Just noticed this is a very old unanswered question!!

Alan May 18, 2014