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You are watching: Does 1/sqrt(n) converge
To identify the convergence or aberration of the collection `sum_(n=1)^oo (-1)^n/sqrt(n)` , us may use the source Test.
In source test, we recognize the border as:
`lim_(n-gtoo) root(n)(|a_n|)= L`
or
`lim_(n-gtoo) |a_n|^(1/n)= L`
Then ,we follow the conditions:
a) `L lt1` then the series converges absolutely
b) `Lgt1` climate the...
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To recognize the convergence or divergence of the collection `sum_(n=1)^oo (-1)^n/sqrt(n)` , we may use the root Test.
In source test, we determine the border as:
`lim_(n-gtoo) root(n)(|a_n|)= L`
or
`lim_(n-gtoo) |a_n|^(1/n)= L`
Then ,we monitor the conditions:
a) `L lt1` then the collection converges absolutely
b) `Lgt1` climate the series diverges
c) `L=1` or does not exist climate the check is inconclusive.The collection may be divergent, conditionally convergent, or for sure convergent.
For the given series `sum_(n=1)^oo (-1)^n/sqrt(n)` , we have `a_n =(-1)^n/sqrt(n).`
Applying the source test, us set-up the border as:
`lim_(n-gtoo) |(-1)^n/sqrt(n)|^(1/n) =lim_(n-gtoo) (1/sqrt(n))^(1/n) Note: |(-1)^n| = 1`
Apply radical property: `root(n)(x) =x^(1/n) ` and also Law the exponent: `(x/y)^n = x^n/y^n` .
`lim_(n-gtoo) (1/sqrt(n))^(1/n) =lim_(n-gtoo) (1/n^(1/2))^(1/n)`
` =lim_(n-gtoo) 1^(1/n) /n^(1/2*1/n)`
` =lim_(n-gtoo) 1^(1/n) /n^(1/(2n))`
` =lim_(n-gtoo) 1 /n^(1/(2n))`
Apply the border property: `lim_(x-gta)<(f(x))/(g(x))> =(lim_(x-gta) f(x))/(lim_(x-gta) g(x))` .
`lim_(n-gtoo) 1 /n^(1/(2n)) =(lim_(n-gtoo) 1 )/(lim_(n-gtoo)n^(1/(2n)))`
` = 1/1`
` =1`
The limit worth `L = 1` implies that the series might be divergent, conditionally convergent, or for sure convergent.
To verify, we use alternating series test on `sum a_n` .
`a_n = 1/sqrt(n)` is positive and decreasing from `N=1`
`lim_(n-gtoo)1/sqrt(n) = 1/oo = 1`
Based on alternating collection test condition, the collection `sum_(n=1)^oo (-1)^n/sqrt(n)` converges.
Apply p-series test on `sum |a_n|` .
`sum_(n=1)^oo |(-1)^n/sqrt(n)|=sum_(n=1)^oo 1/sqrt(n).`
`=sum_(n=1)^oo 1/n^(1/2)`
Based on p-series check condition, we have `p=1/2` the satisfies `0ltplt=1` .
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Thus, the series `sum_(n=1)^oo |(-1)^n/sqrt(n)|` diverges.
Conclusion:
`sum_(n=1)^oo (-1)^n/sqrt(n)` is conditionally convergent due to the fact that `sum_(n=1)^oo (-1)^n/sqrt(n)` is convergent and also `sum_(n=1)^oo |(-1)^n/sqrt(n)|` is divergent.