1)-4n where n=-2 to infinite
2)1/n^2-1 where n=2 to infinite
please help me...i will be thankful :)
2)1/n^2-1 where n=2 to infinite
please help me...i will be thankful :)
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1) This is clearly divergent
.
2) Write the term in partial fraction
1/(n^2-1)= ½[1/(n-1) - 1/(n+1)]
Now sum from n=2 to n=N
½[1/(n-1) - 1/(n+1)]=½[1/1 - 1/2] +½[1/2 - 1/3] +½[1/3 - 1/4]+....+½[1/(N-1) - 1/(N+1)]
If you remove the brackets all terms cancel except
½(1/1) - ½(1/(N+1)
and letting N ->∞ gives
½
so the series converges to ½.
.
2) Write the term in partial fraction
1/(n^2-1)= ½[1/(n-1) - 1/(n+1)]
Now sum from n=2 to n=N
½[1/(n-1) - 1/(n+1)]=½[1/1 - 1/2] +½[1/2 - 1/3] +½[1/3 - 1/4]+....+½[1/(N-1) - 1/(N+1)]
If you remove the brackets all terms cancel except
½(1/1) - ½(1/(N+1)
and letting N ->∞ gives
½
so the series converges to ½.
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step by step:
1. diverges
http://symbolab.com/math/solver/step-by-…
2. converges to 3/4
http://symbolab.com/math/solver/step-by-…
hope this helps
1. diverges
http://symbolab.com/math/solver/step-by-…
2. converges to 3/4
http://symbolab.com/math/solver/step-by-…
hope this helps