Can anyone please explain/show me how to find the next 3 fractions in each of the series below. Thanks.

∞

∑

N=1 1/5n-1= ¼+1/9+1/14+1/19+….

∞

∑

N=1 1/n^2= ¼+1/9+1/14+1/19+….

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- October 2nd 2005, 01:52 PMviolabarNext in series
Can anyone please explain/show me how to find the next 3 fractions in each of the series below. Thanks.

∞

∑

N=1 1/5n-1= ¼+1/9+1/14+1/19+….

∞

∑

N=1 1/n^2= ¼+1/9+1/14+1/19+…. - November 2nd 2005, 05:06 AMbenseries
what your doing here is calculating the formula for integer values from n=1 to infinity and adding them all up. In both the series you have reached n=4 and you want sums for n=5,6, and 7, so just plug these numbers into the 1/5n - 1 and 1/n^2 to find the next 3 terms in the series i.e for n=5

1/5n - 1 = 1/(5*5)-1 = 1/24.....6 and 7 are 1/29 and 1/34

You need to calculate for all values of n again for the 1/n^2 formula as the series you currently have is incorrect

Hope this helps - November 4th 2005, 06:52 PMviolabar
Thank you. It does help. Barbara :)