Question

https://api.agilixbuzz.com/Resz/~Ey6YBAAAAAQuMTRzdauTtA.f73ecUiY1RrHOg6WCBh4bA/19809088,B84/Assets/assessmentimages/45815.jpg
https://api.agilixbuzz.com/Resz/~Ey6YBAAAAAQuMTRzdauTtA.f73ecUiY1RrHOg6WCBh4bA/19809088,B84/Assets/assessmentimages/45715.jpg
https://api.agilixbuzz.com/Resz/~Ey6YBAAAAAQuMTRzdauTtA.f73ecUiY1RrHOg6WCBh4bA/19809088,B84/Assets/assessmentimages/45915.jpg
https://api.agilixbuzz.com/Resz/~Ey6YBAAAAAQuMTRzdauTtA.f73ecUiY1RrHOg6WCBh4bA/19809088,B84/Assets/assessmentimages/46013.jpg

Answer 1

which series will converge?

Answer 2

@Michele_Laino

Answer 3

hint:
the series
\[\huge \sum {k \cdot {n^2},\quad k \in \mathbb{R}} \]
diverges

Answer 4

also the series
\[\huge \sum {a \cdot n,\quad a \in \mathbb{R}} \]

Answer 5

so i would say the second one

Answer 6

it is very easy to check such statements

Answer 7

the second one, is divergent. Here is why:
\[\Large \sum {\frac{1}{2} \cdot n = \frac{1}{2}\left( {1 + 2 + 3 + ...} \right) = + \infty } \]

Answer 8

would it be the third one?
I am kinda confused

Answer 9

yes! The series:
\[\huge \sum\limits_1^{ + \infty } {\frac{1}{{{n^2}}}} \] is convergent

Answer 10

so this one?
https://api.agilixbuzz.com/Resz/~Ey6YBAAAAAQuMTRzdauTtA.f73ecUiY1RrHOg6WCBh4bA/19809088,B84/Assets/assessmentimages/45915.jpg

Answer 11

no, sorry, I meant this:
https://api.agilixbuzz.com/Resz/~Ey6YBAAAAAQuMTRzdauTtA.f73ecUiY1RrHOg6WCBh4bA/19809088,B84/Assets/assessmentimages/45715.jpg

Answer 12

ok thats what i meant by the second one

Answer 13

yes!

Answer 14

Thank you again!

Answer 15

:)