# Homework Help

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The Integral test is applicable if `f` is positive and decreasing function on infinite interval `[k, oo)` where `kgt= 1` and `a_n=f(x)` . Then the series `sum_(n=k)^oo a_n` converges if and only if...

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• Math
The integral test is applicable if `f` is positive and decreasing function on infinite interval `[k, oo) ` where` kgt= 1` and `a_n=f(x)` . Then the series `sum_(n=k)^oo a_n ` converges if and only...

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• Math
`sum_(n=1)^ooarctan(n)/(n^2+1)` Integral test is applicable if f is positive, continuous and decreasing function on infinite interval `[k,oo)` where `k>=1` and `a_n=f(x)` . Then the series...

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• Math
`1/4+2/7+3/12+.....+n/(n^2+3)+........` We can write the series as `sum_(n=1)^oon/(n^2+3)` The integral test is applicable if f is positive , continuous and decreasing function on infinite interval...

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• Math
For the series: `ln(2)/sqrt(2) + ln(3)/sqrt(3)+ ln(4)/sqrt(4)+ ln(5)/sqrt(5)+ ln(6)/sqrt(6) +...`, it follows the formula `sum_(n=2)^oo ln(n)/sqrt(n)` where `a_n = ln(n)/sqrt(n)` . To confirm if...

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• Math
Integral test is applicable if f is positive, continuous and decreasing function on infinite interval `[k,oo)` where `k>=1` and `f(n)=a_n` . Then the series `sum_(n=k)^ooa_n` converges or...

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• Math
`1/3+1/5+1/7+1/9+1/11+..........` The series can be written as, `1/(2*1+1)+1/(2*2+1)+1/(2*3+1)+1/(2*4+1)+1/(2*5+1)+.........` Based on the above pattern we can write the series as,...

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• Math
`1/2+1/5+1/10+1/17+1/26+............` The series can be written as, `1/(1^2+1)+1/(2^2+1)+1/(3^2+1)+1/(4^2+1)+1/(5^2+1)+..........` So based on the above pattern we can write the series as,...

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• Math
Integral test is applicable if `f ` is positive and decreasing function on interval `[k,oo)` where `a_n = f(x).` If `int_k^oo f(x) dx` is convergent then the series `sum_(n=k)^oo a_n` is also...

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• Math
Recall the Integral test is applicable if `f ` is positive and decreasing function on interval `[k,oo)` where `a_n = f(x)` . If `int_k^oo f(x) dx` is convergent then the series `sum_(n=k)^oo a_n`...

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• Math
Integral test is applicable if `f` is positive and decreasing function on interval `[k,oo)` where `a_n = f(x)` . If the integral `int_k^oo f(x) dx` is convergent then the series `sum_(n=k)^oo a_n`...

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• Math
Recall the Integral test is applicable if `f` is positive and decreasing function on interval `[k,oo)` where `kgt=1` and `a_n = f(x)` . If `int_k^oo f(x) dx` is convergent then the series...

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• Math
`sum_(n=1)^oo2/(3n+5)` The Integral test is applicable if f is positive, continuous and decreasing function on the infinite interval `[k,oo)` where `k>=1` and `a_n=f(x)` . Then the series...

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• Math
`sum_(n=1)^oo1/(n+3)` The integral test is applicable if f is positive, continuous and decreasing function on infinite interval `[k,oo)` where `k>=1` and `a_n=f(x)` . Then the series...

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• Math
`sum_(n=0)^oo(-1)^nx^(2n)` `=sum_(n=0)^oo(-x^2)^n` It's a geometric series with ratio`r=-x^2` A geometric series `sum_(n=0)^ooar^n` with ratio r diverges if `|r|>=1` . If `0<|r|<1` , then...

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• Math
`sum_(n=0)^oo(-1)^nx^n` `=sum_(n=0)^oo(-1x)^n` It is a geometric series with common ratio`r=-x` ,so the series converges for `|r|<1` `|-x|<1` `=>|x|<1` `=>-1<x<1` In this...

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• Math
`sum_(n=0)^oo5((x-2)/3)^n` It is a geometric series with first term `a=5` and common ratio `r=(x-2)/3` , so the series converges for `|r|<1` `|(x-2)/3|<1` `=>-1<(x-2)/3<1`...

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• Math
`sum_(n=1)^oo(x-1)^n` It is a geometric series with first term`a=(x-1)` and common ratio `r=(x-1)` , so the series converges for `|r|<1` `|x-1|<1` `=>-1<(x-1)<1` `=>0<x<2`...

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• Math
`sum_(n=0)^oo(2/x)^n` It is a geometric series with common ration`r = 2/x` , so the series converges for `|r|<1` `|2/x|<1` `=>2<|x|` `=>x>2` or `x<-2` Sum of the...

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• Math
`sum_(n=1)^oo(3x)^n` It is a geometric series with common ratio`r=3x` , so the series converges for `|r|<1` `|3x|<1` `=>|x|<1/3` `=>-1/3<x<1/3` In this case, sum of the...

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• Math
`sum_(n=1)^ooarctan(n)` Use the n'th term test for divergence, If `lim_(n->oo)a_n!=0` , then `sum_(n=1)^ooa_n` diverges `a_n=arctan(n)` `lim_(n->oo)arctan(n)=pi/2!=0` As per the divergence...

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• Math
`sum_(n=1)^ooe^(-n)` The series can be expressed as, `=sum_(n=1)^oo1/e^n` `=sum_(n=1)^oo(1/e)^n` It's a geometric series with first term `a=1/e` and ratio `r=1/e` , since `|r|=1/e<1` , the...

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• Math
Recall that the Divergence test follows the condition: If `lim_(n-gtoo)a_n!=0` then sum `a_n` diverges. For the given series `sum_(n=1)^oo ln(1/n)` , we have `a_n = ln(1/n)` . To evaluate it...

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• Math
`sum_(n=2)^oo n/(ln (n))` To determine if the series is convergent or divergent, apply the nth-Term Test for Divergence. It states that if the limit of `a_n` is not zero, or does not exist, then...

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• Math
Recall that infinite series converges to single finite value `S` if the limit if the partial sum `S_n` as n approaches `oo` converges to `S` . We follow it in a formula: `lim_(n-gtoo)...

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• Math
`sum_(n=1)^oo 3^n/n^3` To determine if the series is convergent or divergent, apply the ratio test. The formula for the ratio test is: `L = lim_(n->oo) |a_(n+1)/a_n|` If L<1, the series...

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• Math
`sum_(n=1)^oo(1/(n+1)-1/(n+2))` `S_n=(1/(1+1)-1/(1+2))+(1/(2+1)-1/(2+2))+(1/(3+1)-1/(3+2))+........+(1/(n-1+1)-1/(n-1+2))+(1/(n+1)-1/(n+2))`...

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• Math
`sum_(n=1)^oo(1/n-1/(n+2))` `S_n=(1/1-1/(1+2))+(1/2-1/(2+2))+(1/3-1/(3+2))+..........+(1/(n-1)-1/(n+1))+(1/n-1/(n+2))` `S_n=(1-1/3)+(1/2-1/4)+(1/3-1/5)+.......+(1/(n-1)-1/(n+1))+(1/n-1/(n+2))` This...

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• Math
`sum_(n=1)^oo(4n+1)/(3n-1)` For the given series `a_n=(4n+1)/(3n-1)` `a_n=(4+1/n)/(3-1/n)` `lim_(n->oo)a_n=lim_(n->oo)(4+1/n)/(3-1/n)` `=4/3!=0` As per the n'th term test for divergence, If...

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• Math
`sum_(n=0)^oo(n+10)/(10n+1)` For the series `a_n=(n+10)/(10n+1)` `a_n=(1+10/n)/(10+1/n)` `lim_(n->oo)a_n=lim_(n->oo)(1+10/n)/(10+1/n)` `=1/10!=0` As per the n'th term test for divergence, If...

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• Math
Recall that infinite series converge to a single finite value `S ` if the limit of the partial sum `S_n` as n approaches `oo` converges to `S` . We follow it in a formula: `lim_(n-gtoo)...

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• Math
`sum_(n=0)^oo(1.075)^n` It's a geometric series with ratio `r=1.075` A geometric series with ratio r diverges if `|r|>=1` For the given series `r=1.075>1` So the series diverges.

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• Math
`sum_(n=1)^oo1/(9n^2+3n-2)` Let's rewrite the n'th term of the sequence as, `a_n=1/(9n^2+3n-2)` `=1/(9n^2+6n-3n-2)` `=1/(3n(3n+2)-1(3n+2))` `=1/((3n+2)(3n-1))` Now let's carry out partial fraction...

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• Math
`sum_(n=1)^oo(sin1)^n` Let's find the value of `sin1` from a calculator. `sin(1)=0.84147<1` The series is geometric with first term `a=sin(1)` and common ratio `r=sin(1)` Geometric series...

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• Math
`sum_(n=0)^oo((0.3)^n+(0.8)^n)` `=sum_(n=0)^oo(0.3)^n+sum_(n=0)^oo(0.8)^n` Now both of the above are geometric series having first term 1 and common ratios `0.3,0.8` respectively. Geometric series...

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• Math
`sum_(n=0)^oo(1/2^n-1/3^n)` `=sum_(n=0)^oo1/2^n-sum_(n=0)^oo1/3^n` `=(1/2^0+1/2^1+1/2^2+.......+1/2^oo)-(1/3^0+1/3^1+1/3^2+......+1/3^oo)`...

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• Math
`9-3+1-1/3+........` Let's check the common ratio of the terms, `r=a_2/a_1=-3/9=-1/3` `r=a_3/a_2=1/-3=-1/3` So this is a geometric sequence with common ratio `-1/3` `S_oo=a_1/(1-r)`...

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• Math
`8+6+9/2+27/8+........` Let's find the common ratio of the terms: `r=a_2/a_1=6/8=3/4` `r=a_3/a_2=(9/2)/6=9/12=3/4` So this is a geometric sequence with common ratio of `3/4` `S_oo=a/(1-r)` where a...

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• Math
`sum_(n=1)^oo1/((2n+1)(2n+3))` Using partial fractions, we can write the n'th term of the sequence as, `a_n=1/(2(2n+1))-1/(2(2n+3))` Now the n'th partial sum is,...

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• Math
`sum_(n=1)^oo4/(n(n+2))` Using partial fractions we can write the n'th term of the series as, `a_n=4(1/(2n)-1/(2(n+2)))` `a_n=(2/n-2/(n+2))` Now the n'th partial sum is,...

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• Math
By the definition, the sum of a series is the limit of its partial sums (if it exists). For this series the `N`-th partial sum is `sum_(n=0)^N (-1/5)^n.` This sum is the sum of finite geometric...

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• Math
`sum_(n=0)^oo5(2/3)^n` `=5((2/3)^0+(2/3)^1+(2/3)^2+.........+(2/3)^oo)` `=5(1+(2/3)^1+(2/3)^2+........(2/3)^oo)` This is a geometric series with first term `a=1` and common ratio `r=2/3`...

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• Math
`sum_(n=1)^oo1/(n(n+2))` We can write down the n'th term of the series as, `a_n=1/(n(n+2))` Using partial fractions, `a_n=1/(2n)-1/(2(n+2))` The n'th partial sum of the series is:...

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• Math
`sum_(n=1)^oo1/(n(n+1))` `a_n=1/(n(n+1))` Using partial fractions, we can write the n'th term as, `a_n=1/n-1/(n+1)` The n'th partial sum of the series `S_n` is,...

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• Math
`sum_(n=0)^oo (-0.6)^n` To verify if the series is convergent, apply the Geometric Series Test. Take note that a geometric series has a form `sum` `ar^n` It converges if `|r|lt1` . And it diverges...

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• Math
`sum_(n=0)^oo (0.9)^n` To verify if the series is convergent, apply the Geometric Series Test. Take note that a geometric series has a form `sum` `ar^n` It converges if `|r|lt1` . And it diverges...

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• Math
To verify if the given infinite series: `sum_(n=1)^oo 2(-1/2)^n` converges, recall that infinite series converge to a single finite value S if the limit of the partial sum `S_n ` as n approaches `...

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• Math
`sum_(n=0)^oo (5/6)^n` To verify if the series is convergent, apply the Geometric Series Test. Take note that a geometric series has a form `sum_(n=0)^oo ar^n` or `sum_(n=1)^oo...

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• Math
`sum_(n=1)^oo (n!)/2^n` To verify if the series diverges, apply the ratio test. The formula for the ratio test is: `L = lim_(n->oo) |a_(n+1)/a_n|` If L<1, the series converges. If L>1, the...

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