Criteri di convergenza di serie 2
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
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Exercise:
footnoteN. Rusca Determina la convergenza o la divergenza delle serie seguenti usando il criterio del confronto. abcmulti abc _k^inftyfrac+k^ abc _k^inftyfrack+k^ abc _k^inftyfrack+kk+ abc _k^inftyfrackk+ abc _k^inftyfraclnkk abc _k^inftyfrack^- abcmulti
Solution:
abclist abc Siccome la serie armonica generalizzata per p converge e se diminuiamo il denominatore una frazione aumenta otteniamo: _k^inftyfrac+k^+_k^inftyfrac+k^leq +_k^inftyfrack^infty abc In modo contrario se aumentiamo il denominatore la somma diminuisce dunque _k^inftyfrack+k^ frac+frac+_k^inftyfrack+k^geq frac+frac+_k^inftyfrackk+k^geq frac+frac+_k^inftyfrack+ e siccome l'ultimo termine è la somma armonica da n il tutto diverge a +infty. textbfIn alternativa: _k^inftyfrack+k^ frac+_k^inftyfrack+k^geq frac+_k^inftyfrackk^+k^geq frac+_k^inftyfrack e siccome l'ultimo termine è mezza somma armonica da n il tutto diverge a +infty. abc Diminuiamo il numeratore e otteniamo: _k^inftyfrack+kk+geq _k^inftyfrackkk+geq _k^inftyfrack+ e siccome l'ultimo termine è la somma armonica da n il tutto diverge a +infty. abc In questo caso possiamo usare il criterio di convergenza a infatti lim_kto inftyfrackk+neq dunque la serie non converge! abc Siccome lnkgeq per ke ossia kgeq otteniamo _k^inftyfraclnkkfracln+_k^inftyfraclnkkgeq fracln+_k^inftyfrack e dunque la serie diverge perché è maggiore della serie armonica. abc Sappiamo che k^-underbracek-_geq underbracek^+k+_geq k^geq k^qquad forall kgeq text dunque frack^-leq frack^quad forall kgeq quindi _k^inftyfrack^-leq _k^inftyfrack^infty. abclist
footnoteN. Rusca Determina la convergenza o la divergenza delle serie seguenti usando il criterio del confronto. abcmulti abc _k^inftyfrac+k^ abc _k^inftyfrack+k^ abc _k^inftyfrack+kk+ abc _k^inftyfrackk+ abc _k^inftyfraclnkk abc _k^inftyfrack^- abcmulti
Solution:
abclist abc Siccome la serie armonica generalizzata per p converge e se diminuiamo il denominatore una frazione aumenta otteniamo: _k^inftyfrac+k^+_k^inftyfrac+k^leq +_k^inftyfrack^infty abc In modo contrario se aumentiamo il denominatore la somma diminuisce dunque _k^inftyfrack+k^ frac+frac+_k^inftyfrack+k^geq frac+frac+_k^inftyfrackk+k^geq frac+frac+_k^inftyfrack+ e siccome l'ultimo termine è la somma armonica da n il tutto diverge a +infty. textbfIn alternativa: _k^inftyfrack+k^ frac+_k^inftyfrack+k^geq frac+_k^inftyfrackk^+k^geq frac+_k^inftyfrack e siccome l'ultimo termine è mezza somma armonica da n il tutto diverge a +infty. abc Diminuiamo il numeratore e otteniamo: _k^inftyfrack+kk+geq _k^inftyfrackkk+geq _k^inftyfrack+ e siccome l'ultimo termine è la somma armonica da n il tutto diverge a +infty. abc In questo caso possiamo usare il criterio di convergenza a infatti lim_kto inftyfrackk+neq dunque la serie non converge! abc Siccome lnkgeq per ke ossia kgeq otteniamo _k^inftyfraclnkkfracln+_k^inftyfraclnkkgeq fracln+_k^inftyfrack e dunque la serie diverge perché è maggiore della serie armonica. abc Sappiamo che k^-underbracek-_geq underbracek^+k+_geq k^geq k^qquad forall kgeq text dunque frack^-leq frack^quad forall kgeq quindi _k^inftyfrack^-leq _k^inftyfrack^infty. abclist
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Exercise:
footnoteN. Rusca Determina la convergenza o la divergenza delle serie seguenti usando il criterio del confronto. abcmulti abc _k^inftyfrac+k^ abc _k^inftyfrack+k^ abc _k^inftyfrack+kk+ abc _k^inftyfrackk+ abc _k^inftyfraclnkk abc _k^inftyfrack^- abcmulti
Solution:
abclist abc Siccome la serie armonica generalizzata per p converge e se diminuiamo il denominatore una frazione aumenta otteniamo: _k^inftyfrac+k^+_k^inftyfrac+k^leq +_k^inftyfrack^infty abc In modo contrario se aumentiamo il denominatore la somma diminuisce dunque _k^inftyfrack+k^ frac+frac+_k^inftyfrack+k^geq frac+frac+_k^inftyfrackk+k^geq frac+frac+_k^inftyfrack+ e siccome l'ultimo termine è la somma armonica da n il tutto diverge a +infty. textbfIn alternativa: _k^inftyfrack+k^ frac+_k^inftyfrack+k^geq frac+_k^inftyfrackk^+k^geq frac+_k^inftyfrack e siccome l'ultimo termine è mezza somma armonica da n il tutto diverge a +infty. abc Diminuiamo il numeratore e otteniamo: _k^inftyfrack+kk+geq _k^inftyfrackkk+geq _k^inftyfrack+ e siccome l'ultimo termine è la somma armonica da n il tutto diverge a +infty. abc In questo caso possiamo usare il criterio di convergenza a infatti lim_kto inftyfrackk+neq dunque la serie non converge! abc Siccome lnkgeq per ke ossia kgeq otteniamo _k^inftyfraclnkkfracln+_k^inftyfraclnkkgeq fracln+_k^inftyfrack e dunque la serie diverge perché è maggiore della serie armonica. abc Sappiamo che k^-underbracek-_geq underbracek^+k+_geq k^geq k^qquad forall kgeq text dunque frack^-leq frack^quad forall kgeq quindi _k^inftyfrack^-leq _k^inftyfrack^infty. abclist
footnoteN. Rusca Determina la convergenza o la divergenza delle serie seguenti usando il criterio del confronto. abcmulti abc _k^inftyfrac+k^ abc _k^inftyfrack+k^ abc _k^inftyfrack+kk+ abc _k^inftyfrackk+ abc _k^inftyfraclnkk abc _k^inftyfrack^- abcmulti
Solution:
abclist abc Siccome la serie armonica generalizzata per p converge e se diminuiamo il denominatore una frazione aumenta otteniamo: _k^inftyfrac+k^+_k^inftyfrac+k^leq +_k^inftyfrack^infty abc In modo contrario se aumentiamo il denominatore la somma diminuisce dunque _k^inftyfrack+k^ frac+frac+_k^inftyfrack+k^geq frac+frac+_k^inftyfrackk+k^geq frac+frac+_k^inftyfrack+ e siccome l'ultimo termine è la somma armonica da n il tutto diverge a +infty. textbfIn alternativa: _k^inftyfrack+k^ frac+_k^inftyfrack+k^geq frac+_k^inftyfrackk^+k^geq frac+_k^inftyfrack e siccome l'ultimo termine è mezza somma armonica da n il tutto diverge a +infty. abc Diminuiamo il numeratore e otteniamo: _k^inftyfrack+kk+geq _k^inftyfrackkk+geq _k^inftyfrack+ e siccome l'ultimo termine è la somma armonica da n il tutto diverge a +infty. abc In questo caso possiamo usare il criterio di convergenza a infatti lim_kto inftyfrackk+neq dunque la serie non converge! abc Siccome lnkgeq per ke ossia kgeq otteniamo _k^inftyfraclnkkfracln+_k^inftyfraclnkkgeq fracln+_k^inftyfrack e dunque la serie diverge perché è maggiore della serie armonica. abc Sappiamo che k^-underbracek-_geq underbracek^+k+_geq k^geq k^qquad forall kgeq text dunque frack^-leq frack^quad forall kgeq quindi _k^inftyfrack^-leq _k^inftyfrack^infty. abclist
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