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Series Exercise Converges to 1


Exercise:

Show that, n=14(4n3)(4n+1)=1

Solution:  we have 4(4n3)(4n+1)=A4n3+B4n+1

So 4A+4B=0&A3B=4A=1&B=1

Hence n=14(4n3)(4n+1)=n=114n3n=114n+1=1+n=114(n+1)3n=114n+1

n=14(4n3)(4n+1)=1+n=114n+43n=114n+1=1+n=114n+1n=114n+1=1

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