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| In[1]:= Clear[a, n] In[2]:= SumConvergence[(n^(n - 2))/((a^n)*(n!)), n] Out[2]= Abs[a] > E In[3]:= SumConvergence[(n^(n - 2))/((E^n)*(n!)), n] Out[3]= True All right. So, according to Out[2], the series is convergent if and... |
| Sam, Thanks for responding. This approach is again (I think) using Wolfram Language to supply 'special' functionality. My point in that I would like to get away using Wolfram as, in effect, an API to use when I have complicated Math (or analysis... |
| We're talking about [this pretty complex MIDI file][1], as [seen on YouTube being played by Synthesia][2]. I've downloaded the file and told Mathematica to import it via Import["C:\\Users\\zoga9_000\\Downloads\\Death Waltz.mid"] . Here is what I... |
| If the conditions for which the sum converges are needed, then the GenerateConditions option is handy: Sum[q^k, {k, Infinity}, GenerateConditions -> True] resulting in the output ConditionalExpression[-(q/(-1 + q)), Abs[q] |
| Off-topic reply. Presumably the user of your proposed tutorial is going to have *Mathematica* at his/her disposal. So wouldn't it be much more useful to create the tutorial in the form of a notebook where the user can directly try things, inserting... |