Use the Integral test to determine if the series shown below converges or diverges. Be sure to check that conditions of the integral test are satisfied. Infinity Sigma n =1 for 7/(n^2 +25) I understand that this does converge, but I need help trying to calculate where this converges to

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Use the Integral test to determine if the series shown below converges or diverges. Be sure to check that conditions of the integral test are satisfied. Infinity Sigma n =1 for 7/(n^2 +25) I understand that this does converge, but I need help trying to calculate where this converges to

Calculus1
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|dw:1437979024641:dw|
\[\large \sum\limits_{n=1}^{\infty}~\dfrac{7}{n^2+25}\] like that ?
yes

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Integral test tells us this : If \(\int\limits_1^{\infty} \dfrac{7}{x^2+25}\,dx\) converges, then the given series also converges
so evaluate the integral and see if it converges (finite)
You also have to check the rules right, \[f(x) = \frac{ 7 }{ x^2+25 }\] is continuous on [1, infinity) then f(x) is positive. You also need to check if f is decreasing on [1, infinity) you can do this by taking the derivative, then if all those agree, you must evaluate the integral.
Hint: \[\int\limits \frac{ 1 }{ x^2+a^2 } dx = \frac{ 1 }{ 2 } \tan^{-1}\left( \frac{ x }{ a } \right)+C\]
Ok so I understand that this will converge: so would it be 1/2 tan ^-1 (1/5) then?
yes
ok then would I multiply this by 7 because 7 is the constant?
Correct.
ok I did this and got .69 ( rounded to nearest 100th)... so what would I do next...
No, your result is wrong, it's an improper integral..
\[\int\limits_{0}^{\infty} \frac{ 7 }{ x^2+25 } dx \]
Ok its positive, it is continuous and it is decreasing as we move towards infinity... so:|dw:1437984135224:dw|
|dw:1437984288640:dw|
because the series starts at n =1, I felt a should be = 1 and b = inifinity

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