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rgrewal3

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2. First few terms look like:

(-1)+(-1/2)+(-1/3)+(-1/4)+(-1/5)+(-1/6)+(-1/7)+1/8+1/9+1/10+1/11+1/12+...+1/26+(-1/27)+...+(-1/63)+1/64+...

what's happening is every time you hit a cubic number (8,27,64,125,...) the terms will switch sign until the next cubic number. so let

a1=1+1/2+1/3+1/4+1/5+1/6+1/7

a2=1/8+1/9+...+1/26

a3=1/27+...1/63

etc

your series is then sum((-1)nan)

if an approaches zero (which it does) then the alternating series must converge (alternating series test)

to show the an's converge to zero:

a1<7

a2<19/8

a3<37/27

a4<61/64

.

.

.

an<(3n2+3n+1)/n3

which goes to zero. Done. FYI the numerators are called centred hexagonal numbers

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2. First few terms look like:

(-1)+(-1/2)+(-1/3)+(-1/4)+(-1/5)+(-1/6)+(-1/7)+1/8+1/9+1/10+1/11+1/12+...+1/26+(-1/27)+...+(-1/63)+1/64+...

what's happening is every time you hit a cubic number (8,27,64,125,...) the terms will switch sign until the next cubic number. so let

a1=1+1/2+1/3+1/4+1/5+1/6+1/7

a2=1/8+1/9+...+1/26

a3=1/27+...1/63

etc

your series is then sum((-1)nan)

if an approaches zero (which it does) then the alternating series must converge (alternating series test)

to show the an's converge to zero:

a1<7

a2<19/8

a3<37/27

a4<61/64

.

.

.

an<(3n2+3n+1)/n3

which goes to zero. Done. FYI the numerators are called centred hexagonal numbers

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3. first term is positive, next 3 negative, next 3 positive, next 3 negative, next 4 positive, etc.

the terms of your series can actually be written as (-1)floor(n/pi+.5)|cos(n)|/sqrt(n)

the exponent floor(n/pi+.5) basically counts how many integers are between zeros of the cosine function. so you could leave the first term alone, group the next three negatives together, then the three positives, etc., just as in 2.

it gets a little complicated keeping track of where the "4"s are, but you can see that it will converge quite easily

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