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MathThe 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...

MathThe 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...

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...

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...

MathFor 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...

MathIntegral 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...

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,...

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,...

MathIntegral 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...

MathRecall 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`...

MathIntegral 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`...

MathRecall 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...

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...

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...

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...

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...

Math`sum_(n=0)^oo5((x2)/3)^n` It is a geometric series with first term `a=5` and common ratio `r=(x2)/3` , so the series converges for `r<1` `(x2)/3<1` `=>1<(x2)/3<1`...

Math`sum_(n=1)^oo(x1)^n` It is a geometric series with first term`a=(x1)` and common ratio `r=(x1)` , so the series converges for `r<1` `x1<1` `=>1<(x1)<1` `=>0<x<2`...

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...

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...

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...

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...

MathRecall that the Divergence test follows the condition: If `lim_(ngtoo)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...

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

MathRecall 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_(ngtoo)...

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...

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/(n1+1)1/(n1+2))+(1/(n+1)1/(n+2))`...

Math`sum_(n=1)^oo(1/n1/(n+2))` `S_n=(1/11/(1+2))+(1/21/(2+2))+(1/31/(3+2))+..........+(1/(n1)1/(n+1))+(1/n1/(n+2))` `S_n=(11/3)+(1/21/4)+(1/31/5)+.......+(1/(n1)1/(n+1))+(1/n1/(n+2))` This...

Math`sum_(n=1)^oo(4n+1)/(3n1)` For the given series `a_n=(4n+1)/(3n1)` `a_n=(4+1/n)/(31/n)` `lim_(n>oo)a_n=lim_(n>oo)(4+1/n)/(31/n)` `=4/3!=0` As per the n'th term test for divergence, If...

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...

MathRecall 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_(ngtoo)...

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.

Math`sum_(n=1)^oo1/(9n^2+3n2)` Let's rewrite the n'th term of the sequence as, `a_n=1/(9n^2+3n2)` `=1/(9n^2+6n3n2)` `=1/(3n(3n+2)1(3n+2))` `=1/((3n+2)(3n1))` Now let's carry out partial fraction...

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...

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...

Math`sum_(n=0)^oo(1/2^n1/3^n)` `=sum_(n=0)^oo1/2^nsum_(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)`...

Math`93+11/3+........` 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/(1r)`...

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/(1r)` where a...

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,...

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/n2/(n+2))` Now the n'th partial sum is,...

MathBy 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...

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`...

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:...

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/n1/(n+1)` The n'th partial sum of the series `S_n` is,...

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 `rlt1` . And it diverges...

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 `rlt1` . And it diverges...

MathTo 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 `...

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...

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...

Math`sum_(n=1)^oo(2^n+1)/2^(n+1)` Let's verify it by using n'th term rest for divergence: If `lim_(n>oo)a_n!=0` ,then `sum_(n=1)^ooa_n` diverges `lim_(n>oo)(2^n+1)/2^(n+1)`...